Electricity PDF Past Paper Notes - 2018-2020

UNIT IV: EFFECTS OF CURRENT C H A P TE R 12 ELECTRICITY Syllabus Electric current, potential difference and electric current, Ohm’s law; Resistance, Resistivity, Factors on which the resistance of a conductor depends. Series combination o...

UNIT IV: EFFECTS OF CURRENT C H A P TE R 12 ELECTRICITY Syllabus Electric current, potential difference and electric current, Ohm’s law; Resistance, Resistivity, Factors on which the resistance of a conductor depends. Series combination of resistors, parallel combination of resistors and its application in daily life, Heating effect of electric current and its applications in daily life, Electric power, Interrelation between P, V, I and R. Trend Analysis 2018 2019 2020 List of Concepts OD/D OD D OD D Electric Current, Potential difference, Galvanometer, ammeter, 1 Q (1 M) 2 Q (1 M) 2 Q (3 M) voltmeter Ohm’s Law, Resistance, Resistivity, Factors on which the resistance of a 1 Q (3 M) conductor depends, alloys Series combination of resistors, 1 Q (5 M) 1 Q (3 M) 1 Q (5 M) 1 Q (5 M) 1 Q (1 M) parallel combination of resistors Or 1 Q (5 M) Heating Effect of Electric Current, Or 1 Q (3 M) 1 Q (1 M) Or 1 Q (5 M) 1 Q (5 M) Electric Power. TOPIC - 1 TOPIC - 1 Electric Current, Ohm’s Law Electric Current, Ohm’s Law  Revision Notes TOPIC - 2  Static and Current Electricity: Static electricity deals with the electric charges at rest while the current electricity deals with the electric charges in motion. Resistance in Series and  Electric Current: The electric current is defined as the rate of flow of electric Parallel Combination, charge through any cross section of a conductor. Electric Power and Charge Q Heating Effect Electric current = or I = Time t   Electric current is a scalar quantity.  Ampere: It is the SI unit of current. If one coulomb of charge flows through any cross-section of a conductor in one second, then current through it is said to be one ampere.  Electric circuit: The closed path along which an electric current flows is called an ‘electric circuit’.  Conventional direction of current: Conventionally, the direction of motion of positive charges through the conductor is taken as the direction of current. The direction of conventional current is opposite to that of the negatively charged electrons.  Electrochemical or voltaic cell: It is a device which converts chemical energy into electrical energy.  Galvanometer: It is a device to detect current in an electric circuit.  Ammeter: It is a device to measure current in a circuit. It is a low resistance galvanometer and is always connected in series in a circuit.  Voltmeter: It is a device to measure the potential difference. It is a high resistance galvanometer and is always connected in parallel to the component across which the potential difference is to be measured. Symbol is,  Ohm’s Law: The current through a conductor between two points is directly proportional to the voltage across the two points provided external conditions remain constant. (i) Mathematical expression for Ohm’s law: I ∝ V V = IR (Where R = Resistance) (ii) V-I graph for Ohm’s law:  Resistance (R): It is the property of a conductor to resist the flow of charges through it. (i) S.I. unit of resistance is Ohm (Ω). 1 volt (ii) 1 ohm = 1 ampere  When potential difference is 1 V and current through the circuit is 1 A, then resistance is 1 ohm.  Rheostat: Rheostat is a variable resistor used to regulate current without changing the source of voltage.  Factors on which the Resistance of a Conductor depends: Resistance of a uniform metallic conductor is, (i) Directly proportional to the length of conductor (ii) Inversely proportional to the area of cross-section (iii) Directly proportional to the temperature (iv) Depends on nature of the material.  Resistivity (r): It is defined as the resistance offered by a cube of a material of side 1 m when current flows perpendicular to its opposite faces. (i) Its S.I. unit is ohm-metre (Ωm). (ii) Resistivity does not change with change in length or area of cross-section but it changes with change in temperature. (iii) Range of resistivity of metals and alloys is 10–8 to 10–6 Ωm. (iv) Range of resistivity of insulators is 1012 to 1017 Ωm. (v) Resistivity of alloy is generally higher than that of its constituent metals. (vi) Alloys do not oxidize (burn) readily at high temperature, so they are commonly used in electrical heating devices. (vii) Copper and aluminium are used for electrical transmission lines as they have low resistivity.  Mnemonics Concept: Ohm’s Law Concept: Current Formula Interpretation: Interpretation: To find V=Multiply I and R I=Q/t To find I=Divide V and R Q=I x t To find R=Divide V and I t=Q /I Concept: Connection of ammeter and voltmeter Mnemonics: Am Sleeping Very Patiently Interpretation: Ammeter is connected in Series Voltmeter is connected in Parallel How is it done on the GREENBOARD? Q. Calculate the resistance of a 1 km Step II: r = 1.623 × 10–8 ohm-meter long copper wire of area of corss L = 1 km = 1000 m section 2 × 10–2 cm2. The restivity of A = 2 × 10–2 cm2 = 2 × 10–2 × 10–4 m2 copper is 1.623 × 10–8 ohm-meter. L Solution: Step III: R = ρ A L Step I: R = ρ or 8 R = 1.623  10  1000 A 2  10 2  10 4 where R = Resistance of the wire \ R = 8.1 W L = Length of the were A = Area of cm section of the wire Objective Type Questions 1 mark each (c) 2 A (d) 3 A [OD CBSE, 2020] A Multiple Choice Questions Ans. Correct option: (c) Q. 1. A cylindrical conductor of length ‘l’ and uniform Explanation : Resistivity of the conductor in the first area of cross section ‘A’ has resistance ‘R’. The RA case, ρ =...(i) area of cross section of another conductor of same l material and same resistance but of length ‘2l’ is Resistivity of the conductor in second case, A 3A RA' (a) (b) ρ=...(ii) 2 2 2l  Since, both conductors are of same material and are Explanation : In series connections, the order of at same temperature, so the resistivity of both the elements in the circuit will not affect the amount of conductors will be same. current flowing in the circuit. Therefore, from equations, (i) and (ii), we have : Q. 6. Electrical resistivity of a given metallic wire RA RA' depends upon   (a) its length. (b) its thickness. l 2l ⇒ A’ = 2A (c) its shape. (d) nature of the material. Q. 2. The maximum resistance which can be made using  [NCERT Exemp.] four resistor each of resistance ½Ω is Ans. Correct option : (d) Explanation : The resistivity of a material is constant (a) 2 Ω (b) 1 Ω for a particular material at a constant temperature. (c) 2.5 Ω (d) 8 Ω [OD CBSE, 2020] It only depends on the temperature. Resistivity of Ans. Correct option: (a) material does not depend on length, thickness, and Explanation : Maximum resistance in series = 4 x ½ shape of the material. = 2ohm Q. 7. A current of 1 A is drawn by a filament of an Q. 3. When a 4V battery is connected across an unknown electric bulb. Number of electrons passing through resistor there is a current of 100 mA in the circuit. a cross section of the filament in 16 seconds would The value of the resistance of the resistor is: be where, e = 1.6 × 10 –19 C (a) 1020 (b) 1016 (a) 4 Ω (b) 40 Ω (c) 1018 (d) 1023 (c) 400 Ω (d) 0.4 Ω  [NCERT Exemp.]  [Board SQP, 2020] Ans. Correct option : (a) Ans. Correct option: (b) Explanation : Explanation : V=IR, V = 4 V, I = 100 mA = 0.1 A Given, I =1 Amp. V 4     t =16 s Hence, R= = = 40 Ω. I 0.1 Q ne As I = = Q. 4. Unit of electric power may also be expressed as: t t (a) Volt-ampere (b) Kilowatt-hour   n = (1  T )  1  1619 (c) Watt-second (d) Joule-second e 1.6  10  [Board SQP, 2020]   n = 1020 Ans. Correct Option: (a) Explanation : Unit of electric power is volt-ampere. B Assertions and Reasons Type Questions Q. 5. A cell, a resistor, a key, and ammeter are arranged Directions: In the following questions, a statement as shown in the circuit diagrams of Figure. The of assertion (A) is followed by a statement of current recorded in the ammeter will be: reason (R). Mark the correct choice as: (a) maximum in (i). (a) Both assertion (A) and reason (R) are true and (b) maximum in (ii). reason (R) is the correct explanation of assertion (c) maximum in (iii). (A). (d) the same in all the cases. (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (i) (d) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): A conductor has + 3.2 × 10–19 C charge. Reason (R): Conductor has gained 2 electrons. Ans. Correct option : (c) Explanation : Conductor has positive charge, so it (ii) has lost two electrons. Q. 2. Assertion (A): The resistivity of conductor increases with the increasing of temperature. Reason (R): The resistivity is the reciprocal of the conductivity. Ans. Correct option : (b) (iii) Explanation : The resistivity of the conductors is directly proportional to temperature. Q. 3. Assertion (A): Bending a wire does not affect electrical resistance.  [NCERT Exemp.] Reason (R): Resistance of wire is proportional to Ans. Correct option : (d) resistivity of material.  Ans. Correct option : (a) Ans. Correct option : (a) l Explanation : When two resistances R1 and Explanation : Resistance of wire R = r   R2 connected in parallel than their equivalent  A R1R2 Where r is resistivity of material which does not resistance will be R =. R1 + R2 depend on the geometry of wire. Since when wire is bent its, resistivity, length and area of cross- section do not change, therefore resistance of wire C Very Short Answer Type Questions also remains same. Q. 4. Assertion (A): Two resistance having value R each. Q. 1. Some work is done to move a charge Q from R infinity to a point A in space. The potential of the Their equivalent resistance is. point A is given as V. What is the work done to 2 move this charge from infinity in terms of Q and Reason (R): Given Resistances are connected in V? R [SQP-2020] parallel. Ans. W=QV. 1 Q. 2. State the SI unit of potential difference and name the device used to measure it. A [CBSE Board Delhi, 2019] Topper Answer, 2019 Q. 3. What is the function of a galvanometer in a circuit ? pure metal ? A [CBSE Board Outside Delhi, 2019] R [CBSE Board Delhi- Set- I, 2019] Ans. Due to high resistivity of alloys rather than its Ans. Detect the presence or direction of current. 1 constituting metals. 1  [CBSE Marking Scheme, 2019]  [CBSE Marking Scheme, 2019] Q. 4. Name and define the SI unit of current. Q. 7. Should the resistance of a voltmeter be low or  R [CBSE Board Delhi, Set- II, 2019] high? Give reason.  A [Board Outside Delhi, Set- II, 2019] Ans. Ampere. ½ Flow of 1 coulomb of charge per second. Ans. High. In parallel connection, less current passes 1 coulomb through high resistance. ½+½ 1 ampere = 1 second  ½  [CBSE Marking Scheme, 2019]  [CBSE Marking Scheme, 2019] Q. 8. Name the device that helps to maintain a potential difference across a conductor. Q. 5. Write the function of voltmeter in an electric R [Board Term I, 2016] circuit. R [CBSE Board Delhi, Set-II, 2019] Ans. A battery is used to maintain potential difference Ans. To measure potential difference across two points. across a conductor. 1  [CBSE Marking Scheme, 2019] 1 Q. 9. Write SI unit of resistivity.  R [Board Term I, Set-2, 2015] [DDE 2017] Q. 6. Why are the heating elements of electric toasters Ans. Ohm metre (ohm m). and electric irons made of an alloy rather than a Short Answer Type Questions-I 2 marks each Q. 1. In the experiment to study the dependence of current (I) on the potential difference (V) across a resistor, a student obtained a graph as shown, What does the graph depict about the dependence of current on the potential difference? Find the current that flows through the resistor when the potential difference across it is 2.5V. A [CBSE Board Delhi, 2019]  Topper Answer, 2019 Q. 2. While studying the dependence of potential QM V2 − V2 difference (V) across a resistor on the current Slope = = MP I2 − I2 (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings Value of potential differnce at a point for different values of current and plotted a graph Or R = between V and I. He got a straight line graph Value of current at the same point passing through the origin. What does the straight 1+1=2 line signify? Write the method of determining resistance of the resistor using this graph. Q. 3. The current flowing through a resistor connected [CBSE, Delhi 2019] in a circuit and the potential difference developed Ans. Potential difference (V) is directly proportional to across its ends are as shown in the diagram by current (I) or V ∝ I. 1 milliammeter and voltmeter readings respectively. Method : Finding slope of the graph. 1 (a) What are the least counts of these meters ? [CBSE Marking Scheme 2019] (b) What is the resistance of the resistor ? Detailed Answer: The graph between V and I is a straight line and [CBSE 2019, OD SET-2] passes the origin, this verifies the Ohm’s law. 1 Ans. (a) least count of ammeter = 10 mA ½+½ least count of voltmeter = 0.1 V ½+½ 2.4 (b) = 9.6 ohm (250 mA = 0.25 A) 0.25 The slope gives the resistance of the resistor used in [CBSE Marking Scheme 2019] the circuit. 1 Detailed Answer: (b) Current, I = 250 mA = 250 × 10–3 A (a) Least count of millianmeter = 10 mA Potential difference, V = 2.4 V 1 V 24 Least count of voltmeter = = 0.1 V ½+½ Resistance, R= = = 9.6 W  ½+½ 10 I 250  10 3 Q. 4. The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below: V (volts) 0.5 1.0 1.5 2.0 2.5 3.0 4.0 5.0 I (amepere) 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0 Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.  [CBSE 2018 Delhi Set-1] Topper Answer, 2018 Short Answer Type Questions-II 3 marks each Q. 1. A V-I graph for a nichrome wire is given below. diffidence (V) established across it. This is ohm’s What do you infer from this graph? Draw a law. labelled circuit diagram to obtain such a graph. Resistance of the wire can be calculated as :  [CBSE 2020] V R= I 0.8 = 4 ohm 0.2 This means nichrome wire has a constant value of the resistance 4 ohm. 2 Circuit diagram: Ans. Graph between V and I is a straight line. So, this infers that the flow of current (I) in the conductor is directly proportional to the potential 1  Q. 2. (a) State the relation correlating the electric L current flowing in a conductor and the voltage Ans. Ra = ½ applied across it. Also draw a graph to show this A relationship.  3L  rL Rb = r  =9 = 9 Ra ½ (b) Find the resistance of a conductor if the electric  A / 3  A current flowing through it is 0.35 A when the L/3 1 rL 1 potential difference across it is 1.4 V. [CBSE 2020] Rc = r = = R ½ Ans. (a) The flow of current (I) in the conductor is 3A 9 A 9 a directly proportional to the potential difference Hence Rb > Ra > Rc ½ (V) established across it provided the physical ra = rb = rc because all the three conductors are of conditions remain same. same material. [CBSE Marking Scheme, 2018] 1 Or V = IR Q. 4. What is electrical resistivity ? Derive its SI unit. Graph: In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change ? Justify your answer. U [Delhi Comptt. 31/1, 31/2, 31/3, 2018] Ans. Electrical resistivity of the material of a conductor is the resistance offered by the conductor of length 1 m and area of cross-section 1 m2. 1 RA r = l 2 (a) Given : ohm meter 2 Unit of r = = ohm meter 1 Potential Difference (V)= 1.4 V metre Current (I) = 0.35 A Resistance of wire is doubled if its length is dou- As per formula, V = IR bled. V 1.4 Hence current is reduced to half. So, = = 4ohm 1 I 0.35 100 mA \ Ammeter reading = = 50 mA. 1 2 Q. 3. The figure below shows three cylindrical copper conductors along with their face areas and lengths. [CBSE Marking Scheme, 2018] Compare the resistance and the resistivity of the COMMONLY MADE ERROR three conductors. Justify your answer.  Students often write vague answer. They get confused between the terms resistance and resistivity. ANSWERING TIP  Always write the unit in SI system only. Resistance of wire is directly proportional to length of the wire. So, Resistance of wire get doubled if the length is doubled. AE [SQP - 2018] Q. 5. (a) List the factors on which the resistance of a conductor in the shape of a wire depends. (b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason. (c) Why are alloys commonly used in electrical heating devices ? Give reason. U [Delhi /Outside Delhi 2018] Ans. (a) Factors on which resistance of a conductor depends : (i) Length of conductor [or R ∝ l] (ii) Area of cross-section of the conductor 1 [or R ∝ ] 1 A (b) Metals are good conductor of electricity – as they have low resistivity/have free electrons. Glass is a bad conductor of electricity – as it has high resistivity/have no free electrons. 1 (c) Reason: Alloys have high resistivity/high melting point/alloys do not oxidize (or burn) readily at high tempera- tures. (Any one) 1 [CBSE Marking Scheme, 2018] Detailed Answer: Topper Answer, 2018 Q. 6. State Ohm’s Law. Draw a circuit diagram to Ans. Statement of Ohm’s Law, verify this law indicating the positive and nega- Circuit diagram with polarity of battery, ammeter tive terminals of the battery and the meters. Also and voltmeter show the direction of current in the circuit. Direction of current by arrow. R [Board Term I, 2016] [CBSE Marking Scheme, 2016] 3 Detailed Answer: Ohm’s Law: It states that “Physical conditions ANSWERING TIP remaining same, the current flowing through a conductor is directly proportional to the potential  Candidates should write Ohm’s law difference across its two ends”. correctly and should draw the correct i.e., I ∝V diagram to verify Ohm’s law. V = IR where the constant of proportionality R is called the Q. 7. Calculate the resistance of a 1 km long copper wire electrical resistance. of area of cross section 2 × 10–2 cm2. The resistivity Diagram to verify Ohm’s Law: of copper is 1.623 × 10–8 ohm-meter. A [Board Term I, 2016] l Ans. R = ρ A 1.623 × 10 −8 × 1000 = 2×10 −2×10 −4 m 2 = 0.81 × 10 Ω = 8.1 Ω. 3 Graph : [CBSE Marking Scheme, 2016] COMMONLY MADE ERROR  Sometimes, students fail to write the 2+1 correct formula of resistance. COMMONLY MADE ERROR ANSWERING TIP  Students often write incorrect ohm’s law. While drawing circuit diagram, many of  Practice writing formula in the beginning. them fail to mark the direction of current Make sure you write all essential steps. by arrow. Final answer need to be expressed along with a proper unit. Long Answer Type Questions 5 marks each Q. 1. What does an electric circuit mean ? Name a de- (a) vice that helps to maintain a potential difference across a conductor in a circuit. When do we say (b) that the potential difference across a conductor is 1 volt ?  R [Board Term I, 2016] (iii) Draw a closed circuit diagram consisting of 0.5 m long nichrome wire XY, an ammeter, a voltmeter, Ans. Electric circuit: The closed path along which an four cells of 1.5 V and a plug key. electric current flows is called an ‘electric circuit‘. The device that helps to maintain a potential R [Board Term I, 2016] difference across a conductor in a circuit are— Electric cell, electric battery, electric generator 2+1 Ans. (i) Ammeter The current is said to be one ampere if 1 coulomb of 1 Volt: The potential difference between two points in an electric field is said to be one volt if one joule charge flows through a cross section of conductor of work has to be done in bringing a positive charge per second. of one coulomb from one point to another. (ii) (a) Rheostat (b) Plug key (closed) 1 Joule 1J 1 volt = or 1 V = 2 1 Coulomb 1C Q. 2. (i) Name an instrument that measures electric (iii) current in a circuit. Define unit of electric current. (ii) What are the following symbols mean in an electric circuit. [CBSE Marking Scheme, 2016] 2 + 3  Q. 3. (i) Draw a labelled circuit diagram to study (ii) V = 1.0 V a relationship between potential difference I = 0.25 A (V) across the two ends of a conductor and the current (I) flowing through it. State the formula to V = IR show how I in a conductor varies when V across it V R= is increased step wise. Show this relationship also I on a schematic graph. 1 (ii) Calculate the resistance of a conductor if the = 4 1+2+2 0.25 current flowing through it is 0.25 A when the applied potential difference is 1.0 V. [CBSE Marking Scheme, 2016] U [Board Term I, 2016] Q. 4.  Draw a labelled circuit diagram to study the relationship between the current (I) flowing Ans. (i) through a conductor and the potential difference (V) applied across its two ends. State the formula co-relating the I in a conductor and the V across it. Also show their relationship by drawing a diagram. What would be the resistance of a resistor if the current flowing through it is 0.15 A when the potential difference across it is 1.05 V ? A [Board Term I, 2015] The formula states that the current passing through a conductor is directly proportional to the Ans. potential difference across its ends, provided the physical conditions like temperature, density, etc. remain unchanged. This is Ohm‘s law. V I α V or I = R V = IR. Try Yourself. See Question No. 3(i) of Long Answer Type Question. V I µ V or I = R V = IR. V = 1.05 V I = 0.15 A Putting in equation V = IR or, 1.05 = 0.15 × R 1.05 \ R = =7W 1+2+2 0.15 TOPIC - 2 Resistance in Series and Parallel Combination, Electric Power and Heating Effect Revision Notes  Resistances in series: When two or more resistances are connected end to end so that same current flows through each one of them in turn, they are said to be connected in series. Here, the total resistance is equal to the sum of the individual resistances. Rs = Rl + R2 + R3 +........  Resistances in parallel: When two or more resistances are connected across two points so that each one of them provides a separate path for current, they are said to be connected in parallel. Here, the reciprocal of their combined resistance is equal to the sum of the reciprocals of the individual resistances.  1 1 1 1 = + + +.... RP R1 R 2 R 3  Heating effect of current : When an electric current is passed through a conductor, heat is produced in it. This is known as heating effect of current.  Joule’s law of heating: It states that the heat produced in a conductor is directly proportional to (i) the square of the current I through it, (ii) its resistance R and (iii) the time t, for which current is passed. Mathematically, it can be expressed as : I 2 Rt H = I 2 Rt joule = cal 4.18 VIt Or H = VIt joule = cal 4.18 Practical application of the heating effect of electric current: It is utilised in the electrical   heating appliances such as electric iron, room heaters, water heaters etc. The electric heating is also used to produce light as in an electric-bulb.  Electric energy: It is the total work done in maintaining an electric current in an electric circuit for a given time. Electric energy, W = VIt = I2Rt joule  Electric Fuse: It is a safety device that protects our electrical appliances in case of short circuit or overloading. (i) Fuse is made up of pure tin or alloy of copper and tin. (ii) Fuse is always connected in series with live wire. (iii) Fuse has low melting point. (iv) Current capacity of fuse is slightly higher than that of the appliance.  Electric Power: The rate at which electric energy is consumed or dissipated in an electric circuit : P = VI V2 P = I2R = R  S.I. unit of power = Watt (W) 1 Watt = 1 Volt × 1 ampere  Commercial unit of electric energy = Kilo Watt hour (KWh) 1 kWh = 3.6 × 106 J 1 kWh = 1 unit of electric energy  Electrical power: Electrical power is the rate at which electric energy is consumed by an appliance.  Watt: It is the SI unit of power. The power of an appliance is 1 watt if one ampere of current flows through it on applying a potential difference of 1 volt across its ends. 1 joule 1 watt = = 1 volt × 1 ampere 1 second or 1 W = 1Js–1 = 1VA 1 kilowatt = 1000 W.  Kilowatt hour: It is the commercial unit of electrical energy. One kilowatt hour is the electrical energy consumed by an appliance of 1000 watts when used for one hour. 1 kilowatt hour (kWh) = 3·6 × 106 J  Power rating: The power rating of an appliance is the electric energy consumed per second by the appliance when connected across the marked voltage of the mains.  Efficiency of an electrical device: It is the ratio of the output power to the input power. Output power Efficiency, h = Input power  Mnemonics Concept: Formula of power Mnemonics: Twinkle Twinkle Little Star Power equals I squared R Interpretation: P = I2 R Objective Type Questions 1 mark each 1 A Multiple Choice Questions ⇒R= 25 Ω Q. 1. At the time of short circuit, the electric current in Q. 5. Which of the following represents voltage? the circuit: Work done (a) vary continuously (b) does not change (a) Current ×Time (c) reduces substantially (d) increases heavily [CBSE 2020, Delhi] (b) Work done × Charge Ans. Correct option : (d) Work done × Time (c) Q. 2. Two bulbs of 100 W and 40 W are connected in Current series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be: Work done × Charge (d)  [NCERT Exemp.] (a) 0.4 A (b) 0.6 A Time (c) 0.8 A (d) 1 A [CBSE 2020] Ans. Correct option : (a) Ans. Correct option : (d) Explanation : As we know that, Q. 3. What is the maximum resistance which can be made Work done = Charge × Potential difference 1 using five resistors each of Ω? ⇒Work done = (Current × Time) × Potential 1 5 Difference (a) Ω (b) 10 Ω 5  [Q Charge = Current × time] (c) 5 Ω (d) 1 Ω Workdone  Potential difference   [NCERT Exemp.] Current×Time Ans. Correct option : (d) Q. 6. If the current I through a resistor is increased Explanation : The highest resistance is always given by 100% (assume that temperature remains by connecting the resistors in series. Here, the highest unchanged), the increase in power dissipated will be 1 (a) 100% (b) 200% resistance would be 5 × =1 ohm. 5 (c) 300% (d) 400% Therefore, the maximum resistance is 1 ohm.  [NCERT Exemp.] Q. 4. What is the minimum resistance which can be made Ans. Correct option : (c) 1 Explanation : If I is current and R is resistance then, using five resistors each of Ω? Power, P =I2R 1 5 1 (a) Ω. (b) Ω. Power in first case, P1 = I2R 5 25 100% increase in current means that current becomes 2I   Power in second case, P2 = (2I)2R = 4I2R (c) 1 Ω. (d) 25 Ω.  10 [NCERT Exemp.] Now, increase in dissipated power = P2 – P1 Ans. Correct option : (b) = 4I2R – I2R Explanation : Minimum resistance is obtained when = 3I2R resistors are connected in parallel combination. 3P Percentage increase in dissipated power = 1 × 100 Thus, equivalent resistance obtained by connecting P1 1 = 300% five resistors of resistance Ω each, parallel to Q. 7. In an electrical circuit three incandescent bulbs each other : 5 A, B, and C of rating 40 W, 60 W, and 100 W, re- 1 1 1 1 1 1 1 5 spectively are connected in parallel to an electric = + + + + ⇒ = R 1 1 1 1 1 R 1 source. Which of the following is likely to happen 5 5 5 5 5 5 regarding their brightness? 1 25 ⇒ = (a) Brightness of all the bulbs will be the same. R 1 (b) Brightness of bulb A will be the maximum.  (c) Brightness of bulb B will be more than that of A. Explanation : It is quite clear that in a battery (d) Brightness of bulb C will be less than that of B. circuit, the point of lowest potential is the negative  [NCERT Exemp.] terminal of the battery and the current flows from Ans. Correct option : (c) higher potential to lower potential. Explanation : We know that power is defined as Q. 3. Assertion (A): Electric appliances with metallic rate of doing work. A bulb consumes electric energy body have three connections, whereas an electric and produces heat and light. Now, bulb with more bulb has a two pin connection. power rating will produce more heat and light or Reason (R): Three pin connections reduce heating we can say that power rating of bulb is directly of connecting wires. proportional to the brightness produced by bulb. Ans. Correct option : (c) Therefore, brightness of bulb B with power rating Explanation : The metallic body of an electrical 60 W will be more than the brightness of bulb A appliances is connected to the third pin which is having power rating as 40 W. connected to the earth. This is a safety precaution Q. 8. An electric kettle consumes 1 kW of electric power and avoids eventual electric shock. By doing this the when operated at 220 V. A fuse wire of what rating extra charge flowing through the metallic body is must be used for it? passed to earth and avoid shocks. There is nothing (a) 1 A (b) 2 A such as reducing of the heating of connecting wires (c) 4 A (d) 5 A by three pin connections.  [NCERT Exemp.] Q. 4. Assertion (A): The electric bulbs glow immediate- Ans. Correct option : (d) ly when switch is ON. Explanation : Given that, Reason (R): The drift velocity of electrons in a power = P = 1 kW = 1000 W metallic wire is very high. Voltage = V = 220 Ans. Correct option : (a) P 1000 Explanation : In a conductor there are large Now, I = = = 4.5 A numbers of free electrons. When we close the V 220 circuit, the electric field is established instantly with Now rating of fuse wire must be slightly greater the speed of electromagnetic wave which causes than 4.5 A, that is, 5 A. electron drift at every portion of the circuit. Due to which the current is set up in the entire circuit B Assertions and Reasons Type Questions instantly. The current which is set up does not Directions: In the following questions, a statement wait for the electrons flow from one end of the of assertion (A) is followed by a statement of conductor to another end. It is due to this, the bulb glows immediately when switch is ON. reason (R). Mark the correct choice as: Q. 5. Assertion (A): Copper is used to make electric (a) Both assertion (A) and reason (R) are true and wires. reason (R) is the correct explanation of assertion Reason (R): Copper has very low electrical (A). resistance. (b) Both assertion (A) and reason (R) are true but Ans. Correct option : (a) reason (R) is not the correct explanation of Explanation : A low electrical resistance of copper assertion (A). makes it a good electric conductor. So, it is used to make electric wires. (c) Assertion (A) is true but reason (R) is false. Q. 6. Assertion (A): Silver is not used to make electric (d) Assertion (A) is false but reason (R) is true. wires. Q. 1. Assertion (A): Alloys are commonly used in Reason (R): Silver is a bad conductor. electrical heating devices like electric iron and Ans. Correct option : (c) heater. Explanation : Silver is a good conductor of electricity but it is not used to make electric wires because it is Reason (R): Resistivity of an alloy is generally expensive. higher than that of its constituent metals but the alloys have low melting points then their constituent metals. C Very Short Answer Type Questions Ans. Correct option : (c) Q. 1. Two unequal resistances are connected in parallel. If you are not provided with any other parameters Explanation : Alloy are hold for electrical heating (eg. numerical values of I and R), what can be said devices due to their light restivity and high melting about the voltage drop across the two resistors? point compared to constituent metals. Ans. Voltage-drop is same across both. Q. 2. Assertion (A): In a simple battery circuit the point of Q. 2. What is meant by the statement. “The resistance of lowest potential is positive terminal of the battery. a conductor is one ohm” ?  Reason (R): The current flows towards the point U [CBSE 2020,OD,Set-3] of the lower potential as it flows in such a circuit Ans. The resistance of a conductor is said to be 1 ohm from the positive to the negative terminal. if a current of 1 ampere flows through it when the Ans. Correct option : (d) potential difference across it is 1 volt.  Q. 3. Write the mathematical expression for Joule’s law Ans. Cord is made up of copper wire whereas heating of heating. R [CBSE 2020] element is made up of alloy. Ans. Mathematical expression of Joule’s law of heating is  [CBSE Marking Scheme, 2019] 1 : H = I2Rt Where, H = Produced Heat Detailed Answer: I = Current flowing through the device The cord of an electric oven is usually made of copper or aluminium whose resistance is very low t = Time of current flow so it does not glow. Whereas, its heating element is R = Resistance of the appliance made up of alloy which has very high resistance. Q. 4. What does the cord of an electric oven not glow So, when current is passed through the heating while its heating element does ? element it becomes very hot and glows red.  A [Board Outside Delhi, Set- III, 2019] Short Answer Type Questions-I 2 marks each Q. 1. A student has two resistors- 2 Ω and 3 Ω. She has H=V×Q to put one of them in place of R2 as shown in the H = 40 ×96000 circuit. The current that she needs in the entire H = 384000 J circuit is exactly 9A. Show by calculation which of Q. 3. Define electric power. Write an expression relating the two resistors she should choose electric power, potential difference and resistance. [CBSE 2020 OD] Ans. Electric power : It is the amount of electric energy consumed in a circuit per unit time. V2 Expression : P = R Q. 4. How many 132 Ω resistors in parallel are required [SQP 2020] to carry 5 A on a 220 V line ? Ans. The overall current needed = 9A. [CBSE 2020 OD Set -3] The voltage is 12V Ans. Given V = 220 V, I = 5 A Hence by Ohm’s Law V=IR, V = IR The resistance for the entire circuit = 12/9 = 4/3 Ω. V =R Or R= I R1 and R2 are in parallel. Hence, R=(R1 R2)/(R1 + R2) 4 In parallel combination, let the no. of resistors = x or, = 4R2/(4+R2) = 4/3 132 220 3 = x 5 \ R2 = 2Ω Q. 2. Compute the heat generated while transferring 132 or, = 44 96,000 coulomb of charge in two hours through a x potential difference of 40 V. 132 Ans. (b) Given, Charge (Q)= 96000 C, Time (t) = 2 h, or, x= 44 Potential difference (V) = 40 V Heat generated, H = VIt (where I = Q/t) \ x =3 H=d The number of resistors = 3 Short Answer Type Questions-II 3 marks each Q. 1. Consider the following circuit: What would be the readings of the ammeter and the voltmeter when key is closed ? Give reason to justify your answer. R [CBSE Delhi Comptt. 31/1, 31/2, 31/3, 2018] Ans. R = R1 + R 2 + R3 R = 5 W + 8 W + 12 W = 25 W 1 V =6V V = IR V 6V \ I = = = 0.24 A 1 R 25 W  6 (b) Current in 1st bulb, Hence, Current through 12 W resistance is A P1 100 5 25 I1 = = = A or 0.45 A V 220 11 = 0.24 A Current in 2nd bulb, 6A×12W V = IR = = 2.88 V 1 P2 60 3 25 I2 = = = A or 0.27 A 1+2 V 220 11 Q. 2. Calculate the total cost of running the following Q. 4. Show how would you join three resistors, each electrical devices in the month of September, if the of resistance 9 W so that the equivalent resistance rate of 1 unit of electricity is ` 6.00. of the combination is (i) 13.5 W, (ii) 6 W ? (i) Electric heater of 1000 W for 5 hours daily. U [Delhi /Outside Delhi, 2018] (ii) Electric refrigerator of 400 W for 10 hours daily. Ans. (i) R [CBSE Comptt. 31/1, 31/2, 31/3, 2018] 1000 Ans. P1 = 1000 W = kW, t1 = 5h 1000 400 P2 = 400 W = kW, t2 = 10h 1000 Two 9 ohm resistors in parallel connected to one 9 No. of days, n = 30 ohm resistor in series. E1 =P1 × t1 × n ½ 1 1 1 2 = + = = 1 kW × 5h × 30 = 150 kWh ½ Rp 9 9 9 E2 = P2 × t2 × n 9 400 \ Rp = W = kW × 10 h × 30 2 1000 = 120 kWh ½ 9 R =9W+ W = 13.5 W \ Total energy = (150 + 120) kWh = 270 kWh ½ 2 \ Total cost = 270 × 6 = ` 1620 1 (ii) COMMONLY MADE ERROR  Calculation error should be avoided. Two 9 ohm resistors in series connected to one 9 ANSWERING TIP ohm resistor is parallel Rs = 9 W + 9 W = 18 W 1 1 1 3  While solving numerical, always write = + = formula in the beginning. Keep in mind R 18 9 18 that the essential steps are properly \ R =6W 3 shown and final answer is expressed along with a proper unit. Q. 3. (a) Write Joule’s law of heating. COMMONLY MADE ERROR (b) Two lamps one rated 100 W 220 V, and the other 60 W 220 V, are connected in parallel to electric mains  Students often forget to write formula supply. Find the currents drawn by two bulbs from in the beginning. Keep in mind that the the line, if the supply voltage is 220 V. essential steps are properly shown and U [Delhi /Outside Delhi-2018] final answer is expressed along with a Ans. (a) Joule’s law of heating: Heat produced in a proper unit. resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to the resistance for a given current ANSWERING TIP and (iii) directly proportional to the time for which the current flows through the resistor.  Candidates should do calculations H = I2Rt where, H = Heat produced, I = current, completely otherwise marks are deducted R = Resistance of the conductor and t = Time for for incomplete calculations. which the current flows through the resistor. Detailed Answer: Topper Answer, 2018 Q. 5. Three resistors of 10 Ω, 15 Ω and 20 Ω are connected = 0.2 A in series in a circuit. If the potential drop across the ∴ Current in the circuit = 0.2 A 15 Ω resistor is 3 V, find the current in the circuit ∴ Potential drop across 10 Ω resistor is. and potential drop across the 10 Ω resistor. V = IR = 0.2 A × 10 Ω A [Board Term I, 2016] =2V 1½ + 1½ Ans. In series circuit same current flows through all the [CBSE Marking Scheme, 2016] resistors. Current through 15 Ω resistor, Q. 6. A circuit has a line of 5 A. How many lamps of V 3V 1 rating 40 W, 200 V can simultaneously run on this I = = = R 15 W 5 line safely ? A [Board Term I, 2016] Ans. Given, V = 200 V, P = 40 W, I = 5A, n = ? Q. 8. Give reason for the following: (i) Why are copper and aluminium wires used as nP = VI connecting wires ? VI 200×5 (ii) Why is tungsten used for filament of electric lamps? n= = P 40 (iii) Why is lead-tin alloy used for fuse wires ? AE [Board Term-I- 2015, 2016] 100 = = 25 lamps 4 Ans. (i) These are good conductors of electricity/low resistance, low resistivity. [CBSE Marking Scheme, 2016] 3 (ii) Very high melting point and high resistivity. Q. 7. A bulb is rated at 200V – 40W. What is its (iii) Low melting point. 1+1+1 resistance ? 5 such bulbs are lighted for 5 hours. [CBSE Marking Scheme, 2016] Calculate the electrical energy consumed ? Find Q. 9. Show four different ways in which three resistors the cost if the rate is 5.10 per KWh. of ‘r ’ ohm each may be connected in a circuit. A [Board Term I, 2016] In which case is the equivalent resistance of the Ans. V = 200 V, P = 40 W combination: C [DDE-2015] P = VI (i) Maximum P 40 1 (ii) Minimum [Board Term I, 2014] I = = = A V 200 5 Ans. (a) ½ V 200 R = = I 1 ½ 5 (b) = 200 × 5 = 1000 Ω Total Power = 40 W × 5 = 200 W Time = 5 hrs (c) ½ Electrical energy = 200 W × 5 hrs. = 1000 Wh = 1 KWh. Cost of 1 KWh = ` 5.10 3 (d) ½ \ Total cost = ` 1 × 5.10 = ` 5.10 [CBSE Marking Scheme, 2016] (i) Circuit (a) has maximum resistance ½ (ii) Circuit (b) has minimum resistance ½ Long Answer Type Questions 5 marks each Q. 1. (a) Define Power and state its S1 unit. (b) Calculate the energy consumed by 3 such bulbs (b) A torch bulb is rated 5 V and 500 mA. Calculate: if they glow continuously for 10 hours for complete (i) Power (ii) Resistances (iii) Energy consumed month of November. AE when it is lighted for 2½ hours.R [CBSE 2020 OD] (c) Calculate the total cost if the rate is Rs 6.50 per Ans. (a) Power: It is the amount of electric energy unit. U [CBSE 2020 OD] consumed in a circuit per unit time. Ans. (a) Given, V = 200 volts and P = 100 watt P = W/t V2 V2 (200)2 40000 As P = or R = = = = 400 W Its S.I unit is Watt (W). R R 100 W 100Ω (b) V = 5V 2 I = 500 mA = 0.5 A 2 (b) Electrical energy consumed, E = number of units × (i) P = V × I = 5 × 0.5 = 2.5 W Power of each unit × time × total days V 5 Here, n = 3, P = 100 W, t = 10 hours, Days = 30 (ii) Resistance R = = = 10 ohms I 0.5 So, E = 3 × 100 W × 10 h × 30 = 90,000 Wh = 90 kWh 2 (iii) Energy consumed = P × t (c) Total cost of electricity = Total unit of energy = 2.5 × 2.5 consumed × Cost per unit = 6.25 Wh 3 = 90 kWh × 6.50 = ` 585 1 Q. 2. (a) An electric bulb is rated at 200 V-100 W. What is Q. 3. In the given circuit, A, B, C and D are four lamps its resistance ? U connected with a battery of 60 V. [SQP 2020]  (iv) Find out the total resistance of the circuit R Ans. (i) The lamps are in parallel. 1 (ii) Advantages: If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel. 1 (iii) The lamp with the highest power will glow the brightest. Analyse the circuit to answer the following P=VI In this case, all the bulbs have the same questions. voltage. But lamp C has the highest current. Hence, (i) What kind of combination are the lamps arranged for Lamp C in (series or parallel)? U P = 5 × 60 Watt = 300 W. (the maximum). 1 (ii) Explain with reference to your above answer, what (iv) The total current in the circuit = 3+4+5+3 A = 15A are the advantages (any two) of this combination of lamps? C Voltage = 60V (iii) Explain with proper calculations which lamp V= IR and hence R = V/I glows the brightest? A = 60/15 A = 4A 1 Q. 4. A bulb is rated 40W, 220V. Find the current drawn by it, when it is connected to a 220V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25W, 220V,will be there be any change in the value of current and resistance ? Justify your answer and determine the change. U Ans. [CBSE Delhi, 2019] Topper Answer, 2019  Ans. P = 40 W, V = 220V (b) In an electric circuit two resistors of 12Ω each are P = VI joined in parallel to a 6 V battery. Find the current P 40 2 drawn from the battery. \ I= = = A  A [CBSE Board Delhi, Set- I, 2019] V 220 11 From Ohm’s law, V = IR Ans. V 220 R= = = 1210W I 2 when replaced by 25w, 220 v lamp P 25 5 I= = = A V 220 44 V 220 R= = = 1936W I 5 Low power bulb draw less current since its 1 resistance is higher. Q. 5. (a) With the help of a suitable circuit diagram prove From figure: I = I 1 + I 2 + I3 that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to V V V I1 = , I2 = , I3 = the sum of the reciprocals of the individual resist- R1 R2 R3 ances.  1 1 2  1 1 1  V = + = = 1 /6 ohm ∴ I = V + + = 1 12 12 12  R1 R 2 R 3  R p Rtotal = 6 ohm 1 V 6V 1 1 1 Hence, current I = = =1A ∴ = + + 1 RTotal 6W Rp R1 R 2 R 3 Q. 6. An electric lamp of resistance 20 Ω and a conductor (b) R1 = R2 = 12 Ω and V = 6 V of resistance 4 Ω are connected to a 6 V battery as 1 1 1 1 1 shown in the circuit. Calculate: ∴ = + = + ½ Rp R1 R2 12 12 (a) the total resistance of the circuit. ∴ ½ (b) the current through the circuit. Rp = 6 Ω (c) the potential difference across the (i) electric lamp V 6V and (ii) conductor, and I= 1 = = 1A Rp 6W (d) power of the lamp.  [CBSE Marking Scheme, 2013] Detailed Answer: (a)  U [CBSE Board Delhi, Set- 1, 2019] Ans. (a) R = R1 + R2 1 = 20 Ω + 4 Ω = 24 Ω V (b) I = R 6V = = 0.25 A 1  1+2+2 24 W  Let there be n resistance, each of value (c) (i) For electric lamp: R1, R2.... Rn respectively connected in parallel to V = IR a battery of voltage V. 6 Let current I is sent to the circuit. = × 20 = 5 V 1 If the equivalent resistance is Req, then current 24  V (ii) For Conductor: drawn I = Req V = IR According to the above circuit, 6 = ×4=1V 1 I = I1 + I2 + I3 +.... + In 24  (d) P = VI V V V V V = + + +.... + 6 Req R1 R2 R3 Rn = 5V × A = 1.25 W. 1 24  1 1 1 1 1 [CBSE Marking Scheme, 2019] So, = + + +...... + Req R1 R2 R3 Rn Detailed Answer: Given, Voltage of battery, V = 6 V Therefore, the reciprocal of the equivalent Resistance of electric lamp, R1 = 20 Ω resistance of a group of resistances joined in Resistance of series conductor, R2 = 4 Ω parallel is equal to sum of the reciprocals of (a) Total resistance of circuit, RTotal = R1 + R2 individual resistances. = 20 Ω + 4 Ω (b) In parallel combination, Rtotal is given as = 24 Ω 1 1 1 V = + (b) Using Ohm’s law I = RTotal R1 R2 RTotal  6V (b) Total resistance of the circuit =1 = = 0.25 A 24 W R = R1 + R2 + R3 = 5 + 10 + 15 = 30 ohm (c) Potential difference across Potential difference across the circuit / By (i) Electric lamp, V1 = IR1 ohm’s law = 0.25 A × 20 W V 30 V V = IR or I =   1A 1 = 5V R 30   (ii) Conductor, V2 = IR2 Potential difference across 15 ohm Resistor = = 0.25 A × 4 W 1A × 15 Ω = 15 V = 1V  [CBSE Marking Scheme, 2019] ½ (d) Power of the lamp = I2R = (0.25)2 × 20 W Detailed Answer: = 1.25 W 1+1+2+1 (a) S  uppose the experimental set up comprises of three resistors R1, R2 and R3 of three different values COMMONLY MADE ERROR which are connected in series with an ammeter, key and a battery of known voltage is given as below :  Many students commit errors in substituting the correct values in formulas. ANSWERING TIP  Cross-check the values after substituting them in the formula. Keep in mind, essential steps need to be shown expressed along The key K is closed and the ammeter reading is with a proper unit. recorded. Now, the position of ammeter is changed to anywhere in between the resistors. The ammeter Q. 7. (a) How will you infer with the help of an reading is recorded each time. It is found that there is an identical reading each time, which shows that experiment that the same current flows through same current flows through every part of the circuit every part of a circuit containing three resistors in containing three resistances in series connected to a series connected to a battery ? battery. (b) Consider the given circuit and find the current (b) Req = R1 + R2 + R3 flowing in the circuit and potential difference = 5 Ω + 10 Ω + 15 Ω = 30 Ω across the 15 W resistor when the circuit is closed. V Electric current (I) = Req 30 = = 1A 30 Potential difference across 15W resistor = IR = 1 × 15 = 15 V 3+2 Q. 8. (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected  A [Board Outside Delhi, Set-I, 2019] to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression Ans. (a) (i) Join the three resistors of different values for the equivalent resistance of the combination of in series. the resistors. (ii) Connect them with battery, an ammeter and plug key. (b) Calculate the equivalent resistance of the following (iii) Plug the key and note the ammeter read- network: ing. (iv) Change the position of ammeter to anywhere in between the resistors and note the ammeter reading each time. (v) The ammeter reading will remain same every time. Therefore when resistors are  R [Board Outside Delhi, 2019] connected in series same current flows Ans. (a) Total current I = I1 + I2 + I3 through all resistors, when it is connected to a battery.  Let Rp be the equivalent resistance of R1, Note: If explained with the help of diagram V R2, R3. Then the total current I = 1 give full credit. ½×5 Rp  V2 (iii) P =. R If Reqv is less, power consumed will be more. In the given case, Reqv is lesser in the parallel and thus power consumed will be more.  [CBSE Marking Scheme, 2019] ½ Q. 10. What is meant by electric current ? Name and define S.I. unit. In a conductor electrons are flowing from B to A. What is the direction of conventional  current ? Give justification for your answer. A steady current of 1 Ampere flows through a V V V I1 = , I2 = , I3 =  ½ conductor. Calculate the number of electrons that R1 R2 R3 flow through any section of conductor in 1 second.  1 1 1  V (Charge on electron = 1.6 × 10–19 C) ∴ I = V + + =  ½  R1 R 2 R 3  R p  R [DDE-2014] [Board Term-I, 2015] 1 1 1 1 Ans. Definition of Electric current: Refer quick Review ∴ = + +  ½ S.I. unit of Current: Ampere Rp R1 R 2 R 3 Definition of 1 Ampere: It is the SI unit of current. (b) On applying ohm’s law for each R1, R2, R3 If one coulomb of charge flows through any cross- Rp = 10 ohm section of a conductor in one second, then current 1 1 1 2 1 through it is said to be one ampere. = + = = Direction of Current from A to B - Justification R p 20 20 20 10  1 Justification: Conventionally, the direction of Rp = 10 ohm motion of positive charges through the conductor Equivalent resistance of the network is taken as the direction of current. The direction = Req = R1 + Rp = 10 + 10 = 20 ohm.1 of conventional current is opposite to that of the [CBSE Marking Scheme, 2019] negatively charged electrons. Q. 9. (i) Consider a conductor of resistance ‘R’, length ne I×t 1A × 1s ‘L’, thickness ‘d’ and resistivity ‘r’. I= ⇒n= = = 6.25 × 1018 Now this conductor is cut into four equal parts. t e 1.6 × 10 −19 C What will be the new resistivity of each of these [CBSE Marking Scheme, 2015] 5 parts? Why? Q. 11. Establish a relationship to determine the (ii) Find the resistance if all of these parts are equivalent resistance R of a combination of three connected in: resistors having resistances R1, R2 and R3 connected (a) Parallel in series. Calculate the equivalent resistance of the (b) Series combination of three resistors of 2 Ω, 3 Ω and 6 Ω (iii) Out of the combinations of resistors mentioned joined in parallel. U [Board Term I, 2016, 2015] above in the previous part, for a given voltage Ans.