Electrical Engineering Fundamentals PDF Past Paper 2019-2020

Electrical and Electronic Technical College Department of Computer Technical Engineering Electrical Engineering Fundamental First Class 2019- 2020 First Level ASSISTANT LECTURER OMAR IBRAHIM MUSTAFA 1 Symbols and Abbreviations:- The system of units used in engineering and science is the (International system of units), usually abbreviated to SI units, and is based on the metric system. SI units may be made larger or smaller by using prefixes which denote multiplication or division by a particular amount. Charge: The unit of charge is the coulomb (C) where one coulomb is one ampere second. (1 coulomb = 6.24 x 1018 electrons). 2 Thus, charge, in coulombs 𝑸 = 𝑰𝒕 where I is the current in amperes and t is the time in seconds. Example: - if a current of (5 A) flows for (2 minutes), find the quantity of electricity transferred? Quantity of electricity 𝑄 = 𝐼𝑡 coulombs 𝐼 = 5𝐴, 𝑡 = 2 × 60 = 120𝑠, ℎ𝑒𝑛𝑐𝑒 𝑄 = 5 × 120 = 600𝐶 Force: The unit of forces the Newton (N) where one newton is one-kilogram meter per second squared. Thus, force, in Newton's 𝑭 = 𝒎𝒂 Where m is the mass in kilograms and a is the acceleration in meters per second squared. Gravitational force, or weight, is mg, where 𝒈 = 𝟗. 𝟖𝟏 𝒎/𝒔𝟐. Example: A mass of 5000 g is accelerated at 2 m/s2 by a force. Determine the force needed? Force = mass x acceleration = 5kg x 2m/s2 = 10 kg m/s2 = 10 N. Work: The unit of work or energy is the joule (J) where one joule is one Newton meter. Thus work done on a body, in joules, W=Fs where F is the force in Newton's and s is the distance in meters moved by the body in the direction of the force. Power: The unit of power is the watt (W) where one watt is one joule per second. Power is defined as the rate of doing work or transferring energy. Thus, power, in watts, P = W/t, where W is the work done or energy transferred, in joules, and t is the time, in seconds. Thus, energy, in joules, W=Pt 𝑾 𝒕𝒐𝒕𝒂𝒍 𝒘𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 (𝑱) 𝑱 P= = = 𝒕 𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏 (𝑺) 𝑺 𝑱 = 𝟏 𝒘𝒂𝒕𝒕 = 𝟏 𝑵. 𝒎/𝑺 𝑺 𝑾 𝑸 But V= 𝒂𝒏𝒅 𝑰 = 𝑸 𝒕 𝒘 𝑽.𝑸  P= = = 𝑽. 𝑰 ( 𝒘𝒂𝒕𝒕). But V= I.R (Ohm's law). 𝒕 𝒕 3 𝑽² P = V. I = I². R = 𝒘𝒂𝒕𝒕. 𝑹 Example: A mass of 1000 kg is raised through a height of 10 m in 20 s. What is (a) the work done and (b) the power developed? Example: An electric heater consumes 1.8 M/J when connected to a 250 V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply. 4 Ohm's Law: The ratio of potential difference (V) between any two points on a conductor to the current (I) flowing between them, is constant. Provided the temperature of the conductor does not change. V 𝑉 In other words, = constant OR =𝑅 I 𝐼 5 Resistance in Series: Fig. 1a fig.1b According to ohm's law; V= V1+V2+V3 But 𝑉 = 𝐼𝑅 (Ohm's Law) IR = IR1 + IR2 +IR3 =I (R1 + R2 + R3). R eq =R1+R2+R3 As seen from above, the main characteristics of a series circuit are : 1. Same current flows through all parts of the circuit. 2. Different resistors have their individual voltage drops. 3. Voltage drops are additive. 4. Applied voltage equals the sum of different voltage drops. 5. Resistances are additive. 6. Powers are additive. 6 Voltage Divider Rule: Since in a series circuit, same current flows through each of the given resistors, voltage drop varies directly with its resistance. From fig 2: The total resistance: 𝑹𝒕 = 𝑅1 + 𝑅2 + 𝑅3 = 12 Ω According to Voltage Divider Rule, various voltage drops are: 𝑅1 𝑅2 𝑅3 V1= V. , V2= V. , V3= V. 𝑅1+𝑅2+𝑅3 𝑅1+𝑅2+𝑅3 𝑅1+𝑅2+𝑅3 𝑅1 2 𝑉1 = 𝑉. = 24 × =4𝑉 𝑅𝑡 12 𝑅2 4 𝑉2 = 𝑉. = 24 × =8𝑉 𝑅𝑡 12 𝑅3 6 𝑉3 = 𝑉. = 24 × = 12 𝑉 Fig. 2 𝑅𝑡 12 Example : Using voltage divider rule ( V.D.R. ) ,find V1 , V2 , V3 And V4 from fig. 5. Fig. 5 7 R1 1 V1 = V ------------------- = 21 ------------------- = 3 v R1 + R2 + R3 1+2+4 R2 2 V2 = V -------------------- = 21 -------------- = 6 v R1 + R 2 + R3 1+2+4 R3 4 V3 = V --------------------- = 21 -------------- = 12 v R1 + R2 + R3 1+2+4 R2 + R3 2+4 V4 = V ---------------------- = 21 ------------ = 18 v R1 + R2 + R3 1+2+4 8 Resistances in Parallel: The main characteristics of a parallel circuit are: I. same voltage acts across all parts of the circuit 2. Different resistors have their individual current. 3. Branch currents are additive. Fig.3 4. Conductance are additive. 5. Powers are additive. Three resistances, as joined in Fig. 3 are said to be connected in parallel. In this case (i) p.d. across all resistances is the same (ii) current in each resistor is different and is given by Ohm's Law and (iii) the total current i.. The sum of the three separate currents. 𝑉 𝑉 𝑉 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 , 𝐼= + + 𝑅1 𝑅2 𝑅3 𝑉 Now 𝐼 = Where V is the applied voltage. 𝑅 R = equivalent resistance of the parallel combination 𝑉 𝑉 𝑉 𝑉 1 1 1 1 = + + Or = + + 𝑅𝑒𝑞 𝑅1 𝑅2 𝑅3 𝑅𝑒𝑞 𝑅1 𝑅2 𝑅3 𝑅1𝑅2 R eq = 𝑅1+𝑅2 I Current Divider Rule: 𝑉 𝑉 I I I1 = , I2= 𝑅1 𝑅2 1 I 1R2 V= I. Req. R1 V 1 1 𝑅1𝑅2 1 V=I 1 1 𝑅1+𝑅2 𝑅2 1 1  I1 = I 1 𝑅1+𝑅2 1 𝑅1 2 2 I2 = I 𝑅1+𝑅2 3 3 5 5 4 4 4 4 7 7 8 8 Example : Using current divider rule ( C.D.R. ) , calculate I 1 , I 2 and I 3 from fig. 6. IT=20 Fig. 6 To find I 1 , the other resistances are ( 4 // 10 ). 4 x 10 ----------- = 2.857 Ω 4 + 10 2.857 I 1 = 20 x -------------- = 5.796 A 7 + 2.857 To find I 2 , the other resistances are ( 7 // 10 ). 7 x 10 --------- = 4 Ω 7 + 10 4 I2= 20 x ----------- = 10 A 4+4 To find I 3 , the other resistances are ( 7 // 4 ). 7x4 -------- = 2.545 Ω 7+4 2.545 I 3 = 20 x ------------------ = 4.057 A 2.545 + 10 Example. What is the value of the unknown resistor R in Fig.4 if the voltage drop across the 500 resistor is 2.5 volts? All resistances are in ohm. Fig. 4 Example. Calculate the effective resistance of the following combination of resistances and the voltage drop across each resistance when a P.D. of 60 V is applied between points A and B. Resistance between A&C in fig. 5 is 6║3 = 2 Ω Resistance of branch ACD = 18 + 2 = 20 Ω Now, there are two parallel paths between points A and D of resistances 20 and 5. Fig 5 Hence, resistance between A and D= 20║5 = 4 Ω :. Resistance between A and B= 4 + 8 = 12 Ω 60 Total circuit current = =5𝐴 12 5 Current in branch ACD = 5 × =1𝐴 25.. P.D. across 3 Ω and 6 resistors=1 x 2 =2 V P.D. across 18 Ω resistors = 1 x 18=18 V P.D. across 5 resistors =4 x 5 = 20 V P.D. across 8 resistors = 5 x 8 =40 V Example. Find RAB in the circuit, given in Fig. 9. Fig.9 Ans. 22.5 Open and short circuit in series circuits 1. Open circuit : In this case there are no current flows through the circuit as shown in fig. 1. I=0 Fig. 1 2. Short circuit : If the resistance is short circuited, the current will flow through the short circuit. (No current flows through the shorted resistance) As shown in fig. 2. V I= ------------- R1 + R2 R1 I R2 R3 I Fig.2 Open and short circuit in parallel circuits 1. Open circuit: In this case, there are no current flows in the open branch as shown in fig. 3. Fig. 3 I2=0 I=I1+I3 2. Short circuit: In this case, there is no current flow through R1, R2 and R3 because the total current. (I) pass through the short circuit as shown in fig. 4. Fig. 4 V I = ----------- ri Where ri is the internal resistance of the battery. Kirchhoff's laws 1. Kirchhoff's voltage law (KVL): The algebraic sum of voltages in any closed loop is zero. ∑V=0 Now, from fig. 1, there are three equations according to Kirchhoff's voltage law. V + 1 - 1 + + + E - - - - + 4 V 4 3 Fig. 1 Loop 1: E – V1 – V2 =0 E = V1 + V2 ------- (1) Loop 2: V2 – V3– V4 = 0 V2 = V3 + V4 -------- (2) Loop 3: E – V1 – V3 – V4 = 0 E = V1 + V3 + V4 --------- (3) Example: For the circuit shown in fig. 2, using Kirchhoff's voltage law , find V1 and V2. 2 6V + + + 1 V2 V1 10v - - - Fig. 2 Loop 1: 10 _ V2 = 0 V2 = 10V Loop 2: 10 - 6 - V1 = 0 V1 = 10 – 6 = 4V 2- Kirchhoff's Current Law (KCL): In any electrical network, the algebraic sum of currents meeting at a point (junction) is zero as shown in fig. 3. ∑ I = 0, OR  I in =  I out I5 I1 I4 I3 I2 Fig. 3 I 1 + I 3 = I2 + I 4 + I 5 I 1 + I 3 - I2 - I 4 - I 5 =0 Example : Using Kirchhoff's current law , find I 5 from fig. 4. I2 Fig. 4 At node 1: I1+I2=I3 2+ 3 = 5 A I3=I4+I5 5 =1 +I5 I5=5–1=4A Example : Using Kirchhoff's law , find I 1 , I 2 and I 3 for the circuit shown in fig. 5. Fig.5 I 1 = I 2 + I 3 ---------- ( 1 ) Loop 1: 20 - 5 I 1 - 10 I 2 = 0 20 = 5 I 1 + 10 I 2 /5 I 1 + 2 I 2 = 4 --------- ( 2 ) Loop 2: - 10 I 2 +10 I 3 = 0 I 2 = I 3 -------------- ( 3 ) From Equ. (2 ) I 1 = 4 – 2I 2 ------------- ( 4 ) Sub. Equ. (3) And ( 4 ) in ( 1 ) 4–2I2=I2+I2 4I2=4 I2=1A From Equ. (4) I 1 = 4 – (2 x 1 ) = 2 A I3=I2 I 3 = 1A Delta/Star Transformation: In solving networks (having considerable number of branches) by the application of Kirchhoff's Laws, one sometimes experiences great difficulty due to a large number of simultaneous equations that have to be solved. However, such complicated network can be simplified by successively replacing delta meshes by equivalent star system and vice versa. The two arrangements shown in fig.10 will be electrically equivalent if the resistance as measured between any pair of terminals is the same in both the arrangements. Let us find this condition. Fig.10 𝑅12 𝑅31 𝑅12 𝑅23 𝑅1 = ; 𝑅2 = ; 𝑎𝑛𝑑 𝑅12 + 𝑅23 + 𝑅31 𝑅12 + 𝑅23 + 𝑅31 𝑅23 𝑅31 𝑅3 = 𝑅12 + 𝑅23 + 𝑅31 Note that each resistor of the Y is equal to the product of the resistors in the two closest branches of the divided by the sum of the resistors in the. Star/Delta Transformation: This transformation can be easily done by using equations. 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 𝑅12 = ; 𝑅23 = ; 𝑎𝑛𝑑 𝑅3 𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 𝑅31 = 𝑅2 Note that the value of each resistor of the is equal to the sum of the possible product combinations of the resistances of the Y divided by the resistance of the Y farthest from the resistor to be determined. EXAMPLE Convert the ▲ of Fig. To aY. EXAMPLE. Convert the Y of Fig. to a ▲. Example: Find the input resistance of the circuit between the points A and B of Fig. 11. Solution. For finding RAB we will convert the delta CDE of Fig. 11 (a) into its equivalent star as shown in Fig. 11 (b). 4𝑥8 8𝑥6 4𝑥6 Rcs= = 1.7 𝛺 , REs = = 2.6 𝛺 , RDs = = 1.3 𝛺 18 18 18 6.6𝑥9.3 2.6Ω + 4Ω = 6.6Ω , 1.3Ω +8Ω =9.3Ω, 6.6Ω//9.3Ω = = 3.8 𝛺 6.6+9.3 RT=4Ω+1.7Ω+3.8Ω = 9.5 Ω Example. A bridge network ABCD has arms AB, BC, CD and DA of resistances 1, 1, 2 and 1 ohm respectively. If the detector AC has a resistance of 1 ohm, determine by delta / star transformation, the network resistance as viewed from the battery terminals. Solution. As shown in Fig. 12 (b), delta DAC has been reduced to its equivalent star. Fig. 12 2𝑥1 1 2 RD = = 0.5𝛺, RC = = 0.25𝛺, RA = = 0.5𝛺 2+1+1 4 4 Hence, the original network of Fig. 12 (a) is reduced to the one shown in Fig. 12 (d). As seen, there are two parallel paths between points N and B, one of resistance 1.25Q and the other of resistance 1.5 Q. Their combined resistance is: 0.25Ω+1Ω=1.25Ω, P0.5Ω+1Ω=1.5Ω 1.25Ω // 1.5 Ω = 𝟏.𝟐𝟓 𝒙𝟏.𝟓 𝟏.𝟐𝟓+𝟏.𝟓 = 0.68 Ω Total resistance of the network between points D and B is: RT= 0.5 Ω + 0.68 Ω = 1.18 Ω Example. Use delta-star conversion to find resistance between terminals 'AB' of the circuit shown in Fig.13 (a). All resistances are in ohms. Fig.13 a b c d e Exercises: 1- Find Req for the circuit shown: Ans. = 14.4Ω Ans. = 6Ω Ans.=11.2Ω Ans. =11Ω Thevenins Theorem: The current flowing through load resistance RL connected across any two terminals A and B of a network as shown in fig. 1 is given by: Vth IL =----------------- Rth + RL Where:- Vth is the open circuit voltage across the two terminals A and B where RL is removed. Rth is the internal resistance of the network as viewed back into the network from terminals A and B with voltage source replaced by its internal resistance , while current source replaced by open circuit. RL load resistor Fig. 1 Now, Rth and Vth must be found. Rth could be found as follows: 1. Replace voltage source by short circuit (if there is no internal resistance), while the current source replaced by open circuit. 2. Remove RL from the circuit, then calculate Rth viewed from terminals A and B. Vth could be found as follows: 1. Remove RL and make sure that the voltage or current source is connected. 2. Calculate Vth between points A and B. Example1 : Using Thevenins theorem , find IL in the circuit shown below. Using Thevenins theorem:- 1. Remove RL and Calculate Vth between points A and B. 𝑉 10 VTH = 𝑅3 = 8.2 = 6.77 𝑉 𝑅1+𝑅3 3.9+8.2 RTH:- 1. Replace voltage source by short circuit (if there is no internal resistance), while the current source replaced by open circuit. RTH = R1 // R3 + R2 8.2∗3.9 RTH = + 4.7 = 7.343 𝑘𝛺 8.2+3.9 𝑉𝑇𝐻 6.77 IL = = = 0.636 𝑚𝐴 𝑅𝑇𝐻+𝑅𝐿 7.343+3.3 VL = IL * RL = 0.636mA * 3.3 kΩ = 1.908 V Example 2: Using Thevenins theorem, find current flow R3 in the circuit shown below Sol: find voltage between point A and B. 𝑉1 20 I1 = = = 1𝐴 𝑅1+𝑅2 5 +15 VA,B Left = I1 * R2 = 1*15 = 15V R1=5 R3= 10Ω R5 = 2 I1 Ø ◄ I2 V1= 20V V2= 10V R4 = 8 R2= 15 𝑽𝟐 𝟏𝟎 I2 = = = 𝟏𝑨 𝑹𝟓+𝑹𝟒 𝟐+𝟖 VA.B Right = I 2 * R4 = 1 * 8 = 8V VTH = VA,B L - VA.B R = 15 – 8 = 7V RTH = R1 // R2 + R5 // R4 𝟓∗𝟏𝟓 𝟐∗𝟖 RTH = 𝟓+𝟏𝟓 + = 𝟓. 𝟑𝟓 𝜴 𝟐+𝟖 𝑽𝑻𝑯 𝟕 IL = 𝑹𝑻𝑯+𝑹𝟑 = 𝟓.𝟑𝟓+𝟏𝟎 = 𝟎. 𝟒𝟔 𝑨 EXAMPLE3: - Find the Thevenins equivalent circuit for the network in the shaded area of the network of Fig. 9.27. Then find the current through RL for values of 2 Ω, 10Ω, and 100 Ω. EXAMPLE 4 Find the Thevenins equivalent circuit for the network in the shaded area of the network of Fig. 9.33. NORTON’S THEOREM The theorem states the following: Any two-terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a current source and a parallel resistor, as shown in Fig. The discussion of Thevenins theorem with respect to the equivalent circuit can also be applied to the Norton equivalent circuit. The steps leading to the proper values of IN and RN are now listed. Preliminary: 1. Remove that portion of the network across which the Norton equivalent circuit is found. 2. Mark the terminals of the remaining two-terminal network. RN: 3. Calculate RN by first setting all sources to zero (voltage sources are replaced with short circuits and current sources with open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.) Since RN _ RTh, the procedure and value obtained using the approach described for Thevenins theorem will determine the proper value of RN. IN: 4. Calculate IN by first returning all sources to their original position and then finding the short-circuit current between the marked terminals. It is the same current that would be measured by an ammeter placed between the marked terminals. 5. Draw the Norton equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. The Norton and Thevenins equivalent circuits can also be found from each other by using the source transformation discussed earlier in this chapter and reproduced in Fig. 9.5 1 EXAMPLE:-2 find the Norton equivalent circuit for the network in the shaded area of Fig. Solution: Steps 1 and 2 are shown in Fig. Step 3 is shown in Fig. 3∗6 RN R1 // R2 = =2𝛺 3+6 Step 4 is shown in Fig, clearly indicating that the short-circuit Connection between terminals a and b is in parallel with R2 and eliminates its effect. IN is therefore the same as through R1, and the full battery voltage appears across R1 since V2 = I2 R2 = 0 * 6 =0 V 𝐸 9𝑣 Therefore, IN = = = 3𝐴 𝑅1 3𝛺 Step 5: See Fig. 9.64. This circuit is the same as the first one considered in the development of Thevenins theorem. A simple conversion indicates that the Thevenins circuits are, in fact, the same (Fig.). EXAMPLE:-3 find the Norton equivalent circuit for the network external to the 9-Ω Resistor in Fig. 9.66. Solution: Steps 1 and 2: See Fig. 9.67. Step 3: See Fig. 9.68 RN = R1 + R2 = 5Ω + 4Ω = 9 Ω EXAMPLE:-4 (Two sources) Find the Norton equivalent circuit for the portion of the network to the left of a-b in Fig. 9.71. Solution: Steps 1 and 2: See Fig. 9.72. Step 3 is shown in Fig. 9.73, and Step 4: (Using superposition) for the 7-V battery (Fig. 9.74) Maximum power transfer theorem: A resistor load will abstract maximum power from a network when the load resistance is equal to the resistance of the network as viewed from the output terminals with all voltage sources removed leaving behind their internal resistances and all current sources replaced by open circuit. R L = Rth Vth I L = ----------------- R th + R L V th IL = ----------- 2 R th P = (I L) 2 x R th (V th) 2 (𝑉𝑡ℎ)² P max = ---------------- x Rth , P max = 4 𝑅𝑡ℎ 4 (R th) 2 Loop Current Method (Mesh Analysis) Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit. In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next section, we will consider circuits with current sources. In the mesh analysis of a circuit with n meshes, we take the following three steps. Example1:-For the circuit below, find the branch currents I1, I2, and I3 using mesh analysis. Second method: EXAMPLE2:- Find the current through each branch of the network of Fig. using mesh analysis. Solution: Steps 1 and 2 are as indicated in the circuit. Note that the polarities of the 6 Ω resistors are different for each loop current. Step 3: Kirchhoff’s voltage law is applied around each closed loop in the clockwise direction: EXAMPLE 3:- Find the branch currents of the network of Fig. using mesh analysis. Solution: Steps 1 and 2 are as indicated in the circuit. Step 3: Kirchhoff’s voltage law is applied around each closed loop: Nonlinear direct current circuit: There are, components of electrical circuits which do not obey Ohm's law; that is, their relationship between current and voltage (their I–V curve) is nonlinear. An example is the p-n junction diode (curve at right). As seen in the Fig.1, the current does not increase linearly with applied voltage for a diode. Fig. 1 One can determine a value of current (I) for a given value of applied voltage (V) from the curve, but not from Ohm's law, since the value of "resistance" is not constant as a function of applied voltage. The source-free RC circuit: Consider a series combination of a resistor and an initially charged capacitor, as shown in Fig. 2. Our objective is to determine the circuit response, we assume to be the voltage v(t) across the capacitor. Since the capacitor is initially charged, we can assume that at time t = 0, the initial voltage is 𝑣(0) = 𝑉0 Fig. 2 Applying KCL at the top node of the circuit in Fig. 2, 𝑖𝑐 + 𝑖𝑅 = 0 𝑑𝑣 𝑣 By definition, 𝑖𝑐 = 𝐶 𝑎𝑛𝑑 𝑖𝑅 =. Thus, 𝑑𝑡 𝑅 𝑑𝑣 𝑣 𝐶 + =0 𝑑𝑡 𝑅 Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some external voltage or current source, it is called the natural response of the circuit. The natural response is illustrated graphically in Fig. 3. Note that at t = 0, we have the correct initial condition. As t increases, the voltage decreases toward zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by the lower case Greek letter tau, τ. Fig. 3 This implies that at t = τ, 𝑉0 𝑒 −𝜏⁄𝑅𝐶 = 𝑉0 𝑒 −1 = 0.368 𝑉0 or 𝜏 = 𝑅𝐶 At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants. We can find the current iR(t), 𝑣(𝑡) 𝑉0 −𝑡⁄𝜏 𝑖𝑅 (𝑡) = = 𝑒 𝑅 𝑅 The power dissipated in the resistor is 𝑉02 −2𝑡⁄𝜏 𝑝(𝑡) = 𝑣𝑖𝑅 = 𝑒 𝑅 Example 15: In Fig. 4, let vC(0) = 15 V. Find vC, vx , and ix for t > 0. Fig.4 We find the equivalent resistance. 20 × 5 𝑅𝑒𝑞 = (8 + 12)║5 = = 4Ω 20 + 5 𝜏 = 𝑅𝑒𝑞 𝐶 = 4 × 0.1 = 0.4 𝑠 From Fig. 4, we can use voltage division to get vx; so Practice problem: Refer to the circuit in Fig. 5. Let vC(0) = 30 V. Determine vC, vx , and i0 for t ≥ 0. Fig. 5 Example: The switch in the circuit in Fig. 6 has been closed for a long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy stored in the capacitor. Fig.6 Fig. 7 For t < 0, and using voltage division for fig. 27 (a): 9 𝑣𝑐 (𝑡) = × 20 = 15 𝑉 𝑡 0, the switch is opened, and we have the RC circuit shown in Fig.7 (b) Req = 1 + 9 = 10 Ω The time constant is τ = ReqC = 10 × 20 × 10-3 = 0.2 s Thus, the voltage across the capacitor for t ≥ 0 is v(t) = vC(0)𝑒 −𝑡⁄𝜏 = 15𝑒 −𝑡⁄0.2 V or v(t) = 15e -5t The initial energy stored in the capacitor is The source-free RL circuit: Consider the series connection of a resistor and an inductor, as shown in Fig. 8. Fig. 8 Our goal is to determine the circuit response, which we will assume to be the current i(t) through the inductor. At t = 0, we assume that the inductor has an initial current I0 , or i(0) = I0 Applying KVL around the loop in Fig. 28, vL + vR = 0 This shows that the natural response of the RL circuit is an exponential decay of the initial current as shown in fig.9. Fig. 9 The time constant of RL circuit is, 𝐿 𝜏= 𝑅 Hence 𝑖(𝑡) = 𝐼0 𝑒 −𝑡/𝜏 So we can find the voltage across the resistor as 𝑣𝑅 (𝑡) = 𝑖𝑅 = 𝐼0 𝑒 −𝑡⁄𝜏 The power dissipated in the resistor is 𝑝 = 𝑣𝑅 𝑖 = 𝐼02 𝑅𝑒 −2𝑡⁄𝜏 The energy absorbed by the resistor is 1 𝑤𝑅 (𝑡) = 𝐿𝐼02 (1 − 𝑒 −2𝑡⁄𝜏 ) 2 1 Note that as t →∞, wR (∞) → 𝐿𝐼02 , which is the same as wL(0), the initial energy stored in 2 the inductor as. Again, the energy initially stored in the inductor is eventually dissipated in the resistor. Example: The switch in the circuit of Fig. 10 has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0. Fig.10 When t < 0, resulting circuit is shown in Fig. 11(a). Fig. 11 we combine the 4Ω and 12Ω resistors in parallel to get 4 × 12 = 3Ω 4 + 12 40 Hence 𝑖1 = =8𝐴 2+3 8 × 12 𝑖(𝑡) = =6A t 0, the switch is open and the voltage source is disconnected. We now have the RL circuit in Fig. 11 (b). Combining the resistors, we have 𝑅𝑒𝑞 = (4 + 12)║16 = 8 Ω 𝐿 2 1 𝜏= = = 𝑠 𝑅𝑒𝑞 8 4 Thus 𝑖(𝑡) = 𝑖(0)𝑒 −𝑡⁄𝜏 = 6𝑒 −4𝑡 𝐴 The current after 1/8 s is 1 −4× 𝑖(1/8) = 6𝑒 8 = 3.64 𝐴 Practice problem: For the circuit in Fig. 12, find i(t) for t > 0. Fig.12 Example for Kirchhoff's lows Example 1: For the circuit in Fig. 1, find voltages v1 and v2. From Ohm’s law, 𝑣1 = 2𝑖 , 𝑣2 = 3𝑖 Applying KVL −20 + 2𝑖 + 3𝑖 = 0 5𝑖 = 20 ∴ 𝑖 =4𝐴 Figure 1 𝑣1 = 2 × 4 = 8 𝑉, 𝑣2 = 3 × 4 = 12 𝑉 Example 2: Determine vo and i in the circuit shown in Fig. 2 Applying KVL: −12 + 4𝑖 + 2𝑉𝑜 − 4 + 6𝑖 = 0 𝑉𝑜 = −6𝑖 −12 + 4𝑖 − 12𝑖 − 4 + 6𝑖 = 0 Figure 2 −16 − 2𝑖 = 0 ∴ 𝑖 = −8 𝐴 𝑎𝑛𝑑 𝑉𝑜 = −6 × −8 = 48 V Example 3: Find current io and voltage vo in the circuit shown in Fig.3: Applying KCL to node a, we obtain: 0.5𝑖𝑜 + 3 = 𝑖𝑜 𝑖𝑜 − 0.5𝑖𝑜 = 3 𝑖𝑜 = 6 𝐴, 𝑉𝑜 = 4𝑖𝑜 = 24 𝑉 Figure 3 Example 4: Find the currents and voltages in the circuit shown in Fig. 4 For loop 1 𝑣1 = 8𝑖1 , 𝑣2 = 3𝑖2 , 𝑣3 = 6𝑖3 𝑖1 = 𝑖2 + 𝑖3 …………….(1) −30 + 8𝑖1 + 3𝑖2 = 0 30−3𝑖2 8𝑖1 = 30 − 3𝑖2 , 𝑖1 = ……..(2) 8 For loop 2 Figure 4 𝑣3 − 𝑣2 = 0, ∴ 𝑣3 = 𝑣2 , 6𝑖3 = 3𝑖2 𝑖2 ∴ 𝑖3 = …………..(3), now put (2) and (3) in (1): 2 30 − 3𝑖2 𝑖2 = 𝑖2 + ∴ 𝑖2 = 2 𝐴, 𝑖3 = 1 𝐴 𝑖1 = 2 + 1 = 3𝐴 8 2 𝑣1 = 24 𝑉, 𝑣2 = 6 𝑉 𝑎𝑛𝑑 𝑣3 = 6 𝑉 HW: 1- Find the currents and voltages in the circuit shown in Fig. 5 Figure 5 Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A, i3 =1.25 A. 2- Obtain v1 through v3 in the circuit of Fig. 6 Figure 6 3- Determine v1 through v4 in the circuit in Fig. 7 Figure 7 Superposition Theorem Analyzing a circuit using superposition has one major disadvantage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source. However, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits: Example: 1 Use the superposition theorem to find v in the circuit in Fig. 1. Let 𝑣 = 𝑣1 + 𝑣2 To obtain v1, we set the current source by open as Shown in Fig. 1b and apply KVL: −6 + 8𝑖1 + 4𝑖1 = 0 12𝑖1 = 6; ∴ 𝑖1 = 0.5 𝐴 Figure 1 𝑣1 = 4𝑖1 = 2𝑉 To get v2, we set the voltage source to short circuits, as in Fig. 1c. Using current division: Figure 1 b fig.1C H. W: Using the superposition theorem, find vo in the circuit in Fig. 2. Answer: 12 V Figure 2 Calculate ix of Fig. 3 using superposition. Figure 3. EXAMPLE2: - Using superposition, determine the current through the 4- Ω resistor of Fig. 9.6. Solution: Considering the effects of a 54-V source (Fig. 9.7): RT = R1 + R2 // R3 = 24 + 12 // 4 = 24 +3 = 27 Ω 𝐸1 54 𝑣 I= = = 2𝐴 𝑅𝑇 27 Ω The effect of E1 on the current I3. Using the current divider rule, 𝑅2 𝐼 12𝛺∗2𝐴 24 𝐴 I'3 = = = = 1.5 𝐴 𝑅2+𝑅3 12𝛺+4𝛺 16 Considering the effects of the 48-V source (Fig. 9.8): RT = R3 + R1 // R2 = 4 + 24 //12 = 4 + 8 = 12Ω. 𝐸2 48𝑣 I"3= 𝑅𝑇 = 12 𝛺 =4𝐴 The total current through the 4-Ω resistor (Fig. 9.9) is I3 = I"3 – I'3 = 4A – 1.5 A = 2.5A (direction of I″3)

Electrical Engineering Fundamentals PDF Past Paper 2019-2020

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