Catholic Schools in Ifugao General Physics 2 Midterm PDF 2023-2024

**CATHOLIC SCHOOLS IN IFUGAO** **GENERAL PHYSICS 2** **SECOND SEMESTER -- MIDTERMS A. Y. 2023-2024** **\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_...

**CATHOLIC SCHOOLS IN IFUGAO** **GENERAL PHYSICS 2** **SECOND SEMESTER -- MIDTERMS A. Y. 2023-2024** **\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_** **[LESSON 1]. Electric Charge, Coulomb's Law and Electric Fields** **Topic 1: ELECTROSTATICS** -- the study of all phenomena associated with electric charges at rest **A. CONDUCTIVITY** -- the measure of the ease at which an electric charge moves through a material. - CONDUCTORS -- materials that allow electric charges to freely flow - INSULATORS -- materials that resist the flow of electric charges - SEMICONDUCTORS -- intermediate between a conductor and an insulator DOPING -- the process of adding atoms of different elements to pure semiconductors to improve their conductivity. SUPERCONDUCTOR -- materials that practically offer no resistance to the flow of electric charges below a critical temperature Example. A material that involves hydrogen sulfide which can conduct charges to a temperature of -70^0^C. **B. PROCESS OF CHARGING** The number of protons and electrons in an atom is equal, hence an atom is said to be neutral. However, an atom may gain or lose electrons. If the atom gains electrons, it becomes negatively charged; if it loses electrons, it becomes positively charged. The following discusses how a neutral body or system acquires a positive or a negative charge **1. CHARGING BY FRICTION** -- two neutral bodies are rubbed together. The material that will become positively or negatively charged will depend on its **electron affinity.** - ELECTRON AFFINITY -- measure of the attraction of an atom to an electron, the tendency of an atom to become negatively charged. \* Materials with higher electron affinity are capable of gaining electrons from those of lower electron affinity THE TRIBOELECTRIC SERIES -- a ranking of some materials based on their electron affinity. Materials are arranged in the order of increasing electron affinity from top to bottom. \* In general, when two materials are rubbed together, the one that is higher on the list will become positively charged **2. CHARGING BY CONDUCTION** -- requires physical contact between a charging body and a neutral body. The sign of the charge acquired by the neutral body is the same with that of the charging body. ( when the charging body is positively charged, the neutral body that comes in contact with it will be positively charged, when the charging body is negatively charged, the neutral body that comes in contact with it will acquire a negative charge ). **3. CHARGING BY INDUCTION** -- the process whereby a neutral body becomes charged when it is brought near either a positively charged or negatively charged body. The negative charges on the neutral body are attracted towards the charging body if the latter is positive. They are repelled from the charging body if it is negatively charged. The effect is known as polarization. The neutral body is then grounded either by touching it or using a wire. Earth is huge reservoir of charges. It can donate or accept electrons. The electrons from the neutral body will travel down the ground if the charging body is negative. The electrons will travel up the ground connection to the neutral body if the charging body is positive. The ground connection is removed followed by the charging body, leaving the previously neutral body with a net charge. This net charge is opposite to that of the charging body. **C. THE ELECTROSCOPE** -- a device used to determine the kind of electrical charge a body has. **Example:** The Gold Leaf Electroscope -- composed of a glass case with an insulating stopper through which a metal rod passes. A metal cap is soldered to the top end of the rod. Two thin leaves of gold are attached to the other end of the rod. The glass case is used to protect the leaves from air current and dust. The Electroscope must be given an initial charge before it can be used to determine the charge of another body. The electroscope may be charged either by conduction or induction. Table 1. How an electroscope determines the charge of a body Charge of the Electroscope Effect on the metal leaves Charge of the body ---------------------------- ---------------------------- -------------------- \+ Leaves diverge farther \+ Leaves Collapses \- \- Leaves diverge farther \- Leaves Collapses \+ **==================================================================================================** **FORMATIVE ASSESSMENT 1a: Solve the following problems completely and neatly. 20 points** 1\. Human hair is rubbed against a hard rubber. What charge is acquired by the human hair and hard rubber? Explain in not more than two lines.( 3pts) 2\. A cloth made of silk acquired a charge of magnitude 8.9 nC when it was rubbed with fiber/ryan. a.) Did the cloth loose or gain electrons? (2pts) b.) How many electrons were transferred during the process? (5pts) c.) What is the change in the mass of the cloth made of silk? (5pts) d.) What is the change in the mass of the fiber/ryan? (5pts) **==================================================================================================** **D. COULOMB'S LAW** -- proposed by a French Scientist, Charles-Augustine de Coulomb \- states that the force of attraction between two small charged bodies is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them \- in equation form: F = [\$\\frac{\\text{k\\ }q\_{1\\ }q\_{2}}{r\^{2}}\$]{.math.inline} ; whereF = Force k = constant of proportionality ( 9x 10 N.m^2^/C^2^ ) [*q*~1 ~]{.math.inline}and [*q*~2~]{.math.inline}= electrical charges of body 1 and 2 respectively r = distance between the two charged bodies SUPERPOSITION PRINCIPLE -- states that each charged body will exert a force on another charged body as if no other charges are present **SAMPLE PROBLEMS** 1\. What is the magnitude, direction and nature of the force on a body of charge ^+^4 X 10^-9^ C that is 5 cm away from a second body of charge 5 X 10^-8^ C? ^+^4 X 10^-9^ C 5 X 10^-8^ C C GIVEN: [ *q*~1 ~]{.math.inline}= ^+^4 X 10^-9^ C Required: magnitude and direction of F 5cm or 0.05 m [*q*~2~]{.math.inline} = 5 X 10^-8^ C r = 5 cm or 0.05 m [*q*~1 ~]{.math.inline} [ *q*~2~]{.math.inline} SOLUTION: [Magnitude of the Force ] F = [\$\\frac{\\text{k\\ }q\_{1\\ }q\_{2}}{r\^{2}}\$]{.math.inline} = [\$\\frac{\\left( {9\\ x\\ 10}\^{9}\\ \\frac{\\text{N.m}\^{2}}{C\^{2}} \\right)\\ \\left( {4\\ x\\ 10}\^{- 9\\ }C \\right)\\ \\left( {5\\ x\\ 10}\^{- 8\\ }C \\right)}{{(\\ 0.05\\ m)}\^{2}}\$]{.math.inline} = 7.2 X 10^-4^ N; [directed away from q~1~ since both bodies are] [of the same charge] [Nature of the Force]: Force of Repulsion 2\. Three metallic sphere are arranged in a straight line: 3 [*μ*]{.math.inline}C 8.5 nC 5 [*μ*]{.math.inline}C 2 cm 4 cm q~1~ q~2~ q~3~ Find the magnitude, direction and nature of the resultant electric force acting on q~2~ because of q~1~ and q~3~ Magnitude of the force **First**, solve for the magnitude and direction of the force acting on q~2~ because of q~1~ F~1\ on\ 2~ = [\$\\frac{\\left( {9\\ x\\ 10}\^{9}\\ \\frac{\\text{N.m}\^{2}}{C\^{2}} \\right)\\ \\left( {3\\ x\\ 10}\^{- 6\\ }C \\right)\\ \\left( {8.5\\ x\\ 10}\^{- 9}C \\right)}{{(\\ 0.02\\ m)}\^{2}}\$]{.math.inline} = 5.74 x 10^-1^ N ; q~1~ moves away from q~2~ ( ) **Second**, solve for the magnitude and direction of the force acting on q~2~ because of q~3~ F~3\ on\ 2~ = [\$\\frac{\\left( {9\\ x\\ 10}\^{9}\\ \\frac{\\text{N.m}\^{2}}{C\^{2}} \\right)\\ \\left( {5\\ x\\ 10}\^{- 6\\ }C \\right)\\ \\left( {8.5\\ x\\ 10}\^{- 9}C \\right)}{{(\\ 0.04\\ m)}\^{2}}\$]{.math.inline} = 2.39 x 10^-1^ N; q~3~ is attracted to q~2~ ( ) **Third**, determine the resultant force by getting the sum of F~1\ on\ 2~ and F~3\ on\ 2~ since both are going to the same directions: F~R~ = F~1\ on\ 2~ + F~3\ on\ 2~ = 5.74 x 10^-1^ N + 2.39 x 10^-1^ N = 0.81 N to the left 3. Three identical point charges with charge + 3 X 10^-6^ are placed at each vertex of an equilateral triangle ABC as shown. If the side of the triangle is 0.01 m, find the magnitude and direction of the resultant electric force acting on the charge at vertex A. 4. SOLUTION: **First**, solve for the magnitude and direction of the force acting on q~A~ because of q~B~ F~B\ on\ A~ = [\$\\frac{\\left( {9\\ x\\ 10}\^{9}\\ \\frac{\\text{N.m}\^{2}}{C\^{2}} \\right)\\ \\left( {3\\ x\\ 10}\^{- 6\\ }C \\right)\\ \\left( {3\\ \\ x\\ 10}\^{- 6}C \\right)}{{(\\ 0.01\\ m)}\^{2}}\$]{.math.inline} = 810 N Therefore, q~B~ repels q~A~, thus, F~B\ on\ A~ is directed to the right up at 60^0^ (since the triangle is equilateral) with the + x-axis **Second**, solve for the magnitude and direction of a force acting on q~A~ because of q~C~ F~C\ on\ A~ = [\$\\frac{\\left( {9\\ x\\ 10}\^{9}\\ \\frac{\\text{N.m}\^{2}}{C\^{2}} \\right)\\ \\left( {3\\ x\\ 10}\^{- 6\\ }C \\right)\\ \\left( {3\\ \\ x\\ 10}\^{- 6}C \\right)}{{(\\ 0.01\\ m)}\^{2}}\$]{.math.inline} = 810 N, q~B~ repels q~A~, thus, F~B\ on\ A~ is directed upward to the left at an Angle of 60^0^ (since the triangle is equilateral ) with the - x-axis - To visualize F~B\ on\ A\ and~ F~C\ on\ A~, place q~A~ at the point of origin of the cartesian coordinate plane and solve for the magnitude and direction of the force using the component method of vector addition. +-------------+-------------+-------------+-------------+-------------+ | Y Y | | FORCE | HORIZONTAL | VERTICAL | | | | | COMPONENT | COMPONENT | | 60^0^ 60^0^ | | | | | | | | | | | | X X | | | | | +=============+=============+=============+=============+=============+ | | | F~B\ on\ A~ | \+ 810 N | \+ 810 N | | | | = 810 N | cos 60^0^ = | sin 60^0^ = | | | | | + 405 N | 702 N | +-------------+-------------+-------------+-------------+-------------+ | | | F~C\ on\ A\ | \- 810 N | \+ 810 N | | | | =~ | cos 60^0^ = | sin 60^0^ = | | | | 810 N | - 405 N | 702 N | +-------------+-------------+-------------+-------------+-------------+ | | | Resultant | ∑ F~X~ = 0 | ∑ F~Y~ = 1 | | | | Electric | | 404 N | | | | Force | | | +-------------+-------------+-------------+-------------+-------------+ : Therefore; the F~R~ is equal to 1 404 N, acting upward. **==================================================================================================** **FORMATIVE ASSESSMENT 1b: Solve the following problems completely and neatly. 10 points** 1. What is the magnitude, direction and nature of the force on a body of charge ^+^2.8 X 10^-9^ C that is 5 cm away from a second body of charge ^-^ 7.5 X 10^-8^ C? 2. Three metallic sphere are arranged in a straight line: 3 [*μ*]{.math.inline}C 8.5 nC 5 [*μ*]{.math.inline}C 1.5 cm 3 cm **==================================================================================================** **Capacitor** is an electric component that temporarily stores energy within a circuit. Inside it are two conducting plates facing each other and separated by an insulator referred to as dielectric. This material impedes the continuous passage of electric current through the capacitor and stores it until it is discharged at a later time. The Capacitance (C) of a conductor is defined as the ratio of the charge (Q) on the conductor to the potential (V) produced. C = [\$\\frac{Q}{V}\$]{.math.inline} ; the unit used to measure C is coulomb per volt or farad (f) Example1: Compute the capacitance if the value of the charge stored is 0.3 x 10^-6^ C and the voltage supplied is 1 x 10^3^ V. Given: 0.3 x 10^-6^ C -- Q Unknown: C Solution: C = [\$\\frac{0.3\\ x\\ 10\^{- 6}C}{{1\\ x\\ 10}\^{3}V}\$]{.math.inline} C= 3 x 10^-10^ F 1 x 10^3^ V--- **Factors that affect the capacitance of a capacitor** **Variable** **Effect on Capacitance** ---------------------------------------- --------------------------- ---------- Area of the conducting plates Increase Increase Decrease Decrease Distance between the conducting plates Increase Increase Decrease Decrease Type of dielectric More conducting Decrease Less conducting Increase **Shape of Capacitors** 1. **Parallel-plate capacitors** - two parallel charging plates are separated by a dielectric that contains charges. Its capacitance is directly proportional to the area of the plates as well as to the distance between these plates.  2. **Cylindrical capacitors** - Inner and outer cylindrical structures correspond to the plates of the parallel-plate capacitor. The dielectric is placed between these two charged cylinders. Its capacitance depends on its length. The longer the capacitor provides a higher capacitance, whereas a shorter one provides lower value. Increasing the amount of its dielectric makes a higher capacitance. 3. **Spherical Capacitor** - The internal spherical structure is one of the charged bodies. The other charged body is the outer spherical structure that covers the internal sphere. The dielectric is placed between these two charged spheres. Its capacitance varies with its radius. Increasing its radius boosts the capacitance offered by this type of capacitor **Capacitors in a Circuit** A. Series Connection 1. The total charge stored by a circuit containing the capacitors is of equal amount or constant throughout. 2. The total voltage in the circuit containing the capacitors varies on the amount of voltage across each capacitor. The total voltage is equal to the sum of the individual voltages of the capacitors in the circuit. 3. The reciprocal of the total capacitance due to the capacitors in the circuit is equal to the sum of the individual reciprocals of each capacitance. Note: the total capacitance is lower than the individual capacitances of the capacitors. \ [\$\$\\frac{1}{C\_{\\text{total}}} = \\frac{1}{C\_{1}} + \\frac{1}{C\_{2}} + \\frac{1}{C\_{3}}\\.\\.\\.\\ + \\frac{1}{C\_{n}}\$\$]{.math.display}\ V~1~[\$= \\frac{C\_{1}}{Q\_{1}}\$]{.math.inline} ; V~2~ [\$= \\frac{C\_{2}}{Q\_{2}}\$]{.math.inline} Solution: a. [\$\\frac{1}{C\_{\\text{total}}} = \\frac{1}{C\_{1}} + \\frac{1}{C\_{2}}\$]{.math.inline} \ [\$\$\\frac{1}{C\_{\\text{total}}} = \\frac{1}{\\ {2.5\\ x\\ 10}\^{- 12}F} + \\frac{1}{{4.1\\ x\\ 10}\^{- 12}F}\$\$]{.math.display}\ b. Q~Total~ = C~Total~ V~Total~ c. Q~Total~ = Q~1~ = Q~2~ d. V~1~[\$= \\frac{Q\_{1}}{C\_{1}}\$]{.math.inline} = [\$\\frac{2.325\\ x\\ 10 - 9\\ C}{{2.5\\ x\\ 10}\^{- 12}F}\$]{.math.inline} V~2~[\$= \\frac{C\_{2}}{Q\_{2}}\$]{.math.inline} = [\$\\frac{2.325\\ x\\ 10 - 9\\ C}{{4.1\\ x\\ 10}\^{- 12}\\text{F\\ }}\$]{.math.inline} B. Parallel Connection 1. The total charge is equal to the sum of the individual charges stored in the capacitor in the circuit. 2. The total voltage stored by the circuit containing the capacitors is of equal amount or constant throughout. 3. The total capacitance is equal to the sum of the individual capacitances stored in the capacitors in the circuit. Solution: a. C~Total~ = C~1~+ C~2~ b. Q~Total~ = C~Total~ V~Total~ c. V~Total~ = V~1~=V~2~ d. Q~1~ = C~1~ V~1~ Q~2~ = C~2~ V~2~ **Applications of Capacitance** 1. Charged Parallel-Plate Capacitors - The capacitance in a parallel-plate capacitor is dependent on the area and distance between the charged plates which is mathematically expressed as: \ [∈~0~ = 8.85*x* 10^ − 12^ *F*/*m*]{.math.display}\ C = [4.425 *x*10^ − 11^*F*]{.math.inline} **==================================================================================================** **FORMATIVE ASSESSMENT 1d: Solve the following problems completely and neatly. 15 points** 1. **A capacitor consists of two square metal plates, each measuring 5 x 10 ^-2^m on a side. In between the plates is a sheet of mica measuring 1 x 10 ^-4^ m thick.** a. **What is the capacitance of the capacitor?** b. **If the charge in one plate is 2 x10 ^-\ 8^ C, what is the potential difference between the plates?** 2. **Find the total capacitance for each connection if C~1~ = 8 F , C~2~ = 5 F and C~3~ = 10 F** A. C~1~ C2 C3 B. C~1~ C~2~ C~3~ C. C~1~ C~3~ C~3~ **==================================================================================================** **[LESSON 2: ELECTRODYNAMICS]** **Topic 1: Electric Current** Due to electric potential energy, electrons move from one point to another. Thus, electric potential energy can be transferred to electrons though work done. This movement is possible because of the electric field around the negative charges. The velocity of this motion is known as **drift velocity**. Normally, electrons move to any direction. If this flow is regulated and made to move continuously in one direction, then the flow becomes an **electric current**. As a result, the drift velocity of the moving charges all point to a single direction along the conductor. **Drift velocity and electric current are directly proportional.** This means that a higher drift velocity results in a higher amount of current, and vice versa. Increased repulsion among the electrons that make up a current also causes higher current density ad drift velocity. On the other hand, there is lower current density and lower drift velocity if there is no repulsion that will push the charges away from each other. Mathematically, electric current is computed using the equation *I =* [\$\\frac{q}{t}\$]{.math.inline} The symbol *I* indicates the electric current, and *q* is the amount of charges that pass through a conductor for every unit of time, *t*. The unit for current is coulomb per second (C/s) or ampere (A). This mathematical expression essentially shows that the electric current is directly proportional to the amount of charges that flows at a certain time. More charges flowing through a conductor generate a higher amount of current, whereas less charges result in a lower amount of current. This relationship is further shown in the following examples. *Example* *1a*. Compute the current produced by a +6.5 x 10^18^C chare flowing in 15s. *Solution*: *I =* [\$\\frac{q}{t}\$]{.math.inline} = [\$\\frac{6.5\\ x\\ 10\^{18\\ }\\text{\\ C}}{15\\ s}\$]{.math.inline} = 4.33 x 10^17^ A b\. A steady current of 0.6 A flows through a wire. How much charge passes through the wire in 1 minute? *Solution*: *q = It* = (0.6A) (60s) = 36C **Topic 2: Resistance and Resistivity** An **electrical conductor** is any material that allows the free flow of electric current. A conductor possesses characteristics that either enhance or limit the flow of current passing through it. The limitation to current flow is referred to as **resistance**. Resistance and electric current are inversely proportional. So a greater amount of resistance on a conductor results in a lower amount of current to flow through the conductor. **Table 4.1 Factors that affect resistance and current flow** **Factors** **Effect on Resistance** **Effect on Current Flow** ----------------------------------- -------------------------- ---------------------------- ---------- Electrical resistivity Increase Increase Decrease Decrease Decrease Increase Electrical conductivity Higher Decrease Increase Lower Increase Decrease Temperature Higher Increase Decrease Lower Decrease Increase Length of conductor Longer Increase Decrease Shorter Decrease Increase Cross-sectional area of conductor Higher Decrease Increase Lower Increase Decrease Table 4.1 shows how changing the property of the conductor affects the resistance that it offers. One of the properties included in the table is electrical resistivity. **Electrical resistivity** is an intrinsic property of the material that describes how it resists the electric current flowing through it. Higher electrical resistivity means higher overall resistance of the material, whereas lower resistivity indicates the material's lower resistance. Consequently, current flow is reduced by an increase in the electrical resistivity of the material, whereas a decrease in the resistivity allows more current to flow through the material. The counterpart of electrical resistivity is **electrical conductivity**. As summarized in table 4.1, an increase in the electrical conductivity of the material results in a lower resistance offered by the material increases its resistance and lowers the flow of current through it. The resistivity, length, and cross-sectional area of a conductor can be related to an equivalent resistance through the equation R = [\$\\frac{\\text{ρV\\ \\ \\ \\ \\ }}{A}\$]{.math.inline} In this equation, [*ρ*]{.math.inline} is the resistivity of the conductor, L is its length, A is its cross-sectional area, and R is the equivalent resistance that it can provide. As a constant value, the unit of [*ρ*]{.math.inline} is ohm-meter ([*Ω* − *m*)]{.math.inline}. The unit of resistance is ohm (Ω). *Example a:* Compute the resistance of a conductor given a resistivity of 10.4 [*Ω* − *m*]{.math.inline}, length of 4m, and cross-sectional area of 7.85 x10^-3^ m^2^. *Solution*: R = [\$\\frac{\\text{ρV\\ \\ \\ \\ \\ }}{A}\$]{.math.inline} = [\$\\frac{(10.4\\ \\mathrm{\\Omega} - m)(4m)}{7.85\\ \\ x\\ 10\^{- 3}\\ m\^{2}}\$]{.math.inline} = 5299.36 Ω *Solution*: L = [\$\\frac{\\text{RA}}{\\rho}\$]{.math.inline} = [\$\\frac{(1\\mathrm{\\Omega})(\\pi)(1.295\\ x\\ 10\^{- 3}m)\^{2}}{1.77\\ x\\ 10\^{8}\\mathrm{\\Omega} - m}\$]{.math.inline}= [2.976*x*10^ − 14^*m*]{.math.inline} **Topic 3: Electromotive Force** **Electromotive force** or EMF is not a force. Instead, it is the potential energy given to a unit charge to make it flow through a conductor or around a complete circuit. The EMF acts like a charge pump that causes charges to flow through a circuit. As a measurable quantity, EMF is measured using the unit volt (V). Electromotive force is what the voltage source provides to a circuit. It is the "push" given to the electric charges for them to flow from the source to the components of the circuit. This "push" is provided by the cell or the battery connected to the circuit and is defined beforehand. Without a battery, there would be no EMF that will make the charges flow, and therefore no current. Similar to electromotive force is the **potential difference** (PD) across a circuit. Potential difference is an actual consideration of the potentials in the circuit. The existence of PD also identifies the flow of charges through the circuit. Without this difference, there will be no electric potential, thus making the flow of charges through the conductor impossible. Both the EMF and PD are measured in terms of voltage (V). **Ohm's Law** In 1827, Georg Simon Ohm discovered the relationship among voltage, current, and resistance. HE found out that electricity acts similarly to water pipe. Through this observation, he was able to summarize the relationship among EMF or voltage (V), electric current (I), and resistance (R) through the **Ohms Law**. In equation form, Ohm's Law is stated as follows: V = IR **Example a:** Using Ohm's Law, solve for the electric current of a conductor given a voltage of 25 V and a resistance of 10 Ω. **Solution**: I = V/R = 25V / 10Ω = 2.5A **b:** An electric water heater uses 15A of current when plugged to a 220 V outlet. What is the resistance provided by the appliance? **Solution**: R = V / I = 220V / 15A = 14.67 Ω **Topic 4: Electric Circuits** The current flows along a conductor, where it is brought from its source to where electrical energy is needed such as your appliances. The pathway for the current to move to and from the source and the appliance is called **electric** **circuit**. A functional circuit has to be "closed" or must form a closed loop. **Closed circuits** allow the current to flow from the source of the current to the load where the current is needed. Figure 4.5 shows an example of this type of circuit. Notice that all the components of the circuit are connected by a closed loop. On the other hand, an "open" circuit does not form a closed loop; the resulting circuit would then be non-functional. **Open circuits** have gaps where current cannot flow. Thus, electric current cannot be delivered to the load where it is needed. Figure 4.6 shows an example of an open circuit. Look at the components of the circuit. Notice the gap.  Look at figure 4.7. See the difference between a closed circuit and an open circuit. **Schematic diagrams** make it easy to draw circuits. Table 4.2 presents the basic components in circuits and their schematic representations. Table 4.2 Basic components of schematic diagrams of electric circuits. A **resistor** is an electronic component used to provide a specific amount of resistance. Generally, it can be considered as a load because loads provide resistance to current flow. The components of a circuit may be connected in series or in parallel. Take a look at the differences between these two basic types of connections. **The Series Circuit** In **series circuit,** all components are connected using a single pathway. In other words, a series circuit is characterized by a single loop for current to flow. The current is the same for all the components along this circuit. The *total voltage* is the sum of each individual voltage across the circuit, and the *total resistance* of the circuit is the sum of the individual resistances of each circuit load. However, the PD of the voltage for each individual circuit component is not the same as the total voltage. These relationships are summarized by the following formulas: V~total~ = V~1~ + V~2~ + V~3~ +....... V~n~ I~total~ = I~1~ + I~2~ + I~3~ +.............. I~n~ R~total~ = R~1~ + R~2~ + R~3~ +.............R~n~ **Example 1.** Compute the individual values and the total values of the voltage, the current, and the resistance of the series circuit below. **Solution**: The total resistance in the circuit is computed as follows: R~total~ = R~1~ + R~2~ = 6Ω + 3Ω = 9Ω Because the total voltage is 20V, you can compute the total current as follows: I~total~ = V~total~ / R~total~ = 20V / 9Ω = 2.22A V~2~ = I~2~R~2~ = (2.22A) (3Ω) = 6.66 V **The Parallel Circuit** **Parallel circuits** use branches to allow current to pass through more than one path, unlike in the series circuit. The voltage between two points in the circuit does not depend on the path taken; thus, the individual voltages in a parallel circuit are the same as the total voltage. However, unlike in the series circuit, the current in each load is not the same as the total current in the circuit. The total current is the sum of the individual currents across the resistors. The reciprocal of the total resistance in this type of circuit is equal to the sum of the reciprocals of the individual resistances. Always remember that the total resistance is always less than the individual resistances. Here are the following formulas for a parallel circuit: I~total~ = I~1~ + I~2~ + I~3~ +........... I~n~ [\$\\frac{1}{R\_{\\text{total}}} = \\ \\frac{1}{R\_{1}} + \\ \\frac{1}{R\_{2}} + \\ \\frac{1}{R\_{3}} + \\ldots\\ \\frac{1}{R\_{n}}\$]{.math.inline} **Example:** Compute the individual values and the total values of the voltage, the current, and the resistance of the parallel circuit below. **Solution**: The total resistance of the circuit is computed as follows: [\$\\frac{1}{R\_{\\text{total}}} = \\ \\frac{1}{R\_{1}} + \\ \\frac{1}{R\_{2}} = \\ \\frac{1}{2\\mathrm{\\Omega}} + \\ \\frac{1}{4\\mathrm{\\Omega}} = \\frac{3}{4\\mathrm{\\Omega}}\$]{.math.inline} = 1.33 Ω **Ohmic and Non-Ohmic Materials** Circuit components can be classified as either ohmic or non-ohmic. Ohmic components show the relationship between voltage and the current as in Ohm's law. This means that the way these components behave in a circuit can be predicted using the said law. Furthermore, the important factors to consider for ohmic components as these are placed in a circuit are voltage, current, and resistance. Examples of ohmic components are resistors and ordinary conducting wires. On the other hand, **non-ohmic components** do not behave as ohmic components. Ohm's law does not apply in the way these components operate; thus, other actors are considered when these components are placed in a circuit. Examples of non-ohmic components are bulb filaments and semiconductors such as transistors and diodes. **Usage of Electric Energy** **Electrical Power in an Electric Circuit** Circuits facilitate the delivery of electrical energy through current flow to people who need to do their daily task. As influenced by the EMF, electric current moves in the circuit from a source to electrically powered equipment. The equipment then converts this energy to work or other forms as governed by the *law of conservation of energy.* You learned from your earlier physics classes that the law of conservation of energy states that *energy is neither created nor destroyed: it is just transformed from one form to another.* The rate of this conversion is referred to as **electric power.** Mathematically, electric power is computed using the equation: P = VI In this equation, P represents the electric power delivered, V is the voltage or EMF, and I is the current delivered. A variation can be obtained using Ohm's Law that would involve the internal resistance or R of the equipment that converts electrical energy. P = [\$\\frac{V\^{2}}{R}\$]{.math.inline} This is interpreted as the power lost due to resistance. Electric power is measured in terms of the unit watt (W). Example 1: **a**. An electric water heater consuming a 140-W of power has been connected to a 220-V outlet. How much current is in the heating element? Solution: From the equation for electric power delivered, the amount of current in the heating element is given by the following: I = [\$\\frac{P}{V}\$]{.math.inline} = = [\$\\frac{140\\ W}{220\\ V}\$]{.math.inline} = 0.64 A The current in the heating element of the water heater is approximately 0.64 A. **b**. Suppose you have a flashlight that is supplied with a 0.50-A current with a voltage of 3.0 V. how much electric power is received by the flashlight bulb? Solution: The electric power received by the flashlight bulb is given by the following: P = (3.0 V) (0.5 A) = 1.5 W An electric power of approximately 1.5 W is delivered to the flashlight bulb. **c**. A coil receives an electric power of 4500 W from a supplied voltage of 240V. What is the resistance of the coil? Solution: The resistance of the coil is computed as follows: R = [\$\\frac{{(240\\ V)}\^{2}}{4500W}\$]{.math.inline} = 12.8 Ω The coil has a resistance of approximately 12.8 Ω **Heat Generation from Electric Current** The electrical energy that passed through a current can be converted into heat by employing a circuit component with a higher amount of resistance. The resistance provided by this component impedes the flow of the current and allows the energy it brings to accumulate and be converted into heat. The amount of heat generated per second by the current flow on a device is computed using the equation **Heat (J) generated per second = I^2^ R** In this equation, I represent the current in the circuit and R is the resistance. This indicates that the amount of heat generated by the current flow in a conductor is directly affected by the amount of current that passes through it and also by the amount of resistance offered by the conductor. This is the reason that you have to consider the current requirement in choosing your home appliances. Choosing a home appliance with the least current requirement can help prevent power loss and save on electrical energy. Example 2: a\. An electric water heater with a resistance of 8Ω draws a 15-A current when plugged into an electric outlet. What is the rate of heat generation? Solution: The amount of heat generation by the electric heater is given by the following: Heat generation rate = (15 A)^2^ (8Ω) = 1800 W = 1.8 kW The rate of heat generation by the heater is 1.8 kW. **Physiological Effects of Electric Shock** **Essential Question: How can we reduce electrical hazards at home?** The electric energy delivered through electric circuits can be both beneficial and hazardous. It is beneficial when it is converted into various usable forms of energy for consumption. However, improper use of electricity is dangerous and may even cause death. The table below provides a summary of the possible effects if an electric current passes through the human body. **Physiological Effects of Electric Current on Human Body** ------------------------------------------------------------- ---------------------------------------------------------------- **Amount of Current Flowing in the Body** **Physiological Effect** 0.001 A Mild tingling sensation 0.01A to 0.19A Muscles Spasms 0.2 A Heart fibrillation or heart beating in an uncontrolled manner. Higher than 0.2A Heart stops beating Any electrical device can have any of the values provided in the table. This amount depends on both voltage and resistance by virtue of Ohm's Law as discussed previously. To recap, Ohm's Law is stated as follows: voltage = (current)(resistance). This implies that, I = [\$\\frac{V}{R}\$]{.math.inline} One of the ways of addressing electric hazards is by the efficient electrical grounding of the circuit of an appliance. You learned how to do electrical grounding in a previous module when you studied charging by induction. This process requires a connecting wire from the main circuitry into direct contact with the ground to allow the flow or spill of excess electric charges from the appliance to the ground. **Applications of Circuitry** Circuits provide the pathway from the source of the electrical energy to where it needs to be converted to different forms. Circuitry and electrical energy have varied applications from batteries, to light bulbs, to music players, to videos and gaming. **Batteries and Light Bulbs** The simplest application of circuitry is seen in how a light bulb works. Common laboratory experiments in electricity involve connecting light bulbs to batteries in different connections. **Household wiring** Your household provides you with the best visualization of how electric circuits work in delivering electrical energy. You know which appliances are included in the circuit and how the use of electricity can be optimized. **Selection of Fuses** The flow of current in a conductor generates heat along its path. Too much heat, however, can be dangerous to the overall circuit and may cause fire. They only allow a certain amount of current to pass through them; otherwise, a fuse burns out if there is an excessive amount of current. If the fuse burns out, it will shut down the entire circuitry, thus preventing damage. It is always safer to use small fuse than a large one. By selecting the proper fuse to be used in your circuitry, you can be protected from electrical accidents due to faulty wiring. **Surface Charge on Junction between wires Made of Different Materials** Current flows uniformly along a conductor made of a single material. If a different material is placed at an end of the conductor, the flow of current can be impeded in a way that is similar to a capacitor. The charges that accumulate in the junction between these materials are referred to as **surface charges.** **Fig. 6.5** Surface charges in-between ends of different materials.  **==================================================================================================** **FORMATIVE ASSESSMENT 2. Solve the following problems completely and neatly. Use a one whole sheet of yellow pad paper.** (35pts.) 1\. A current of 3A flows through a lamp for 8 seconds. How much charge has passed through a point in the circuit? (5pts.) 2\. The amount of charge that passes a point in a circuit in 5 seconds is 0.2 C. Calculate the size of the current? (5pts.) 3\. A current of 0.25A flows for 3 minutes. Calculate how much charge passes through the lamp. (5pts.) 4\. Calculate the area of cross-section of a wire if its length is 1.0m, its resistance is 23Ω and the resistivity of the material is 1.84 x 10^-6^ Ω m. (5pts.) 5\. In a circuit, a 2 A is flowing through the bulb. The voltage across the bulb is 16 V. What is the bulbs resistance? (5pts.)\ 6. You light a light bulb with a 10 V battery. If the bulb has a resistance of 5 Ω. How much current is flowing? (5pts.)\ 7. If a circuit has a resistance of 44 Ω and a current 5 A. What is the voltage? (5pts.) **==================================================================================================** **LESSON 3: ELECTROMAGNETISM** **Topic 1: Fundamentals of Magnetism** ***Magnetism*** is the ability of a magnetic material to attract other magnetic materials. A material to attract other magnetic materials. A material possessing this ability is called a magnet. A magnet maybe natural or artificial. The most common natural magnet is lodestone. ***Magnetic materials*** are those attracted by a magnet. Examples of these are iron, nickel, cobalt, and some alloy like steel. Artificial magnets can be made using magnetic materials. Nonmagnetic materials, such as wood, paper, and glass are attracted by magnets and cannot be made into magnets. ***Poles*** are portions in a magnet, usually near its ends, where the magnetic force is greatest. The north pole or the north-seeking pole of the magnet is the end of the magnet that will point to the north when suspended freely. Magnetization is the process of making a material temporarily or permanently magnetic. A magnetic material may be magnetized by (a) stroking it with permanent magnet in one direction, (b) allowing an electric current to pass through it, or (c) induction due to Earth's magnetism. **Types of Magnetic Materials** Magnetic materials are classified based on their reaction to an applied magnetic field. The three classifications are: ferromagnetic, paramagnetic and diamagnetic. Ferromagnetic materials are strongly attracted by a magnet. Some examples include iron, cobalt, and nickel. *Paramagnetic materials* have net magnetic moments due to unpaired electrons. They are weakly attracted to magnets. Transition metals such as palladium, platinum, and actinide elements are paramagnetic. Paramagnetic materials become more magnetic when cooled. *Diamagnetic materials* weakly respond to a magnetic field. Magnetization exists only when an external magnetic field is applied. Metals, nonmetals, water and organic compounds are diamagnetic. **Topic 2: Magnetic Forces and Magnetic Fields** A *magnetic field* is a region of space, which has the ability to exert a magnetic force on magnetic dipoles and on moving electric charges. **Magnetic Force on Moving Charges** The force on charged particle moving with velocity v in a magnetic field is equal to the charges of the particle multiplied to the cross product of its velocity and the magnetic field. In symbols, **F=qv x B** Where F is the force, **q** is the charge, **v** is the velocity of the charge, and B is the magnetic field. The magnitude of the cross product of vectors v and B is given by vB sin[*θ*]{.math.inline} where [*θ*]{.math.inline} is the smaller angle between v and B. Thus, the magnitude of the magnetic field is given by the equation **B=** [\$\\frac{\\mathbf{F}}{\\mathbf{\\text{qv}}\\mathbf{\\sin}\\mathbf{\\theta}}\$]{.math.inline} The SI unit for magnetic field is the tesla, symbolized by an uppercase letter T. **1T= 1 N/(C.m/s)= 1 N/A.m** Another unit commonly used is the gauss, represented by capital letter G. **1G=** [**10**^ **−** **4**^]{.math.inline}**T** **Electromagnetic Induction** The process of producing an induced electromotive force due to a change in magnetic flux is called **ELECTROMAGNETIC INDUCTION**. Faraday conducted several experiments on electromagnetic induction. He moved a wire through the poles of a horseshoe magnets, plunged a bar magnet into and out of a coil, and removed the magnet after it touches the core of the coil. His experiments showed that there is an induced voltage whenever there is a relative motion between the conductor and the magnet. In other words, E is induced only when B is changing. There are two laws that are important in electromagnetic induction -- Faraday's law and Lenz's law. **FARADAY'S LAW** Faraday formulated his law of electromagnetic induction based on the results of his experiments. Faraday's law states that the induced electromotive force in a closed loop is equal to the rate of change of magnetic flux through the loop in symbols, [**ε**]{.math.inline}**= -** [\$\\frac{\\mathbf{\\mathrm{\\Delta}\\phi}}{\\mathbf{\\Delta}\\mathbf{t}}\$]{.math.inline} The negative sign denotes Lenz's law, which will be further discussed in the succeeding section. Since the magnetic flux is equal to BAcosθ, a change in magnetic field, area of the surface or orientation, or angle between B and A constitutes a change in flux. If there are N number of turns, [**ε**]{.math.inline}**= -N** [\$\\frac{\\mathbf{\\mathrm{\\Delta}\\phi}}{\\mathbf{\\Delta}\\mathbf{t}}\$]{.math.inline} Sample Problem A circular loop of wire of radius 0.25 m is exposed to an external magnetic field that is perpendicular to the plane of the loop. For a time, the magnetic field decreases at a rate of 0.025 T/s. What is the magnitude of the induced electromotive force in the loop? Given: r= 0.25m [\$\\frac{\\mathrm{\\Delta}B}{\\mathrm{\\Delta}t}\$]{.math.inline}= - 0.025 T/s (The negative sign denotes that the field is decreasing) Solution: Solve first for the area of the loop. A= [*πr*^2^]{.math.inline} A= (3.14) (o.25m[)^2^]{.math.inline} A= 0.196 [*m*^2^]{.math.inline} Since the magnetic field is perpendicular to the plane of the loop, the angle between the normal to the plane of the loop and the magnetic field is zero. Thus, using [\$\\mathbf{\\varepsilon =}\\frac{\\mathbf{- \\mathrm{\\Delta}\\phi}}{\\mathbf{\\Delta}\\mathbf{t}}\$]{.math.inline} **=** [\$\\frac{\\mathbf{- \\mathrm{\\Delta}(BA\\ cos\\ \\theta)}}{\\mathbf{\\Delta}\\mathbf{t}}\$]{.math.inline}**= - A cos** [**θ**]{.math.inline} **(**[\$\\frac{\\mathbf{\\mathrm{\\Delta}B}}{\\mathbf{\\mathrm{\\Delta}t}}\$]{.math.inline}**) = (-0. 196** [**m**^**2**^]{.math.inline}**) cos 0 ( -0.025 T/s) = 4.9 x** [**10**^ **−** **3**^]{.math.inline}**V.** **LENZ'S LAW** **Lenz's law of electromagnetic induction** states that the direction of the current induced in a conductor by a changing magnetic field (as per [Faraday's law of electromagnetic induction](https://www.electrical4u.com/faraday-law-of-electromagnetic-induction/)) is such that the [magnetic field](https://www.electrical4u.com/magnetic-field/) created by the induced [current](https://www.electrical4u.com/electric-current-and-theory-of-electricity/) ***opposes*** the initial changing magnetic field which produced it. **=================================================================================================** **FORMATIVE ASSESSMENT 3. Solve the following problems completely and neatly. Use a one whole sheet of yellow pad paper. 10 points** 1. At the instant an electron enters a uniform magnetic field of 0.500 T directed along the positive x-axis, its velocity has the following components: [*v*~*x*~]{.math.inline}= 2.00 x [10^6^]{.math.inline}m/s, [*v*~*y*~]{.math.inline}= 0 m/s and [*v*~*z*~]{.math.inline}= 6.00 x [10^6^]{.math.inline}m/s. Find the radius and pitch of the helical path followed by the electron. 2. A circular loop of wire of radius 26 cm is exposed to an external magnetic field that is perpendicular to the plane of the loop. For a time, the magnetic field decreases at a rate of 23 T/s. What is the magnitude of the induced electromotive force in the loop? **==================================================================================================** **LESSON 4.** Theoretical and experimental approaches to solve multi-concept and rich-context problems involving electricity and magnetism **=================================================================================================** **FORMATIVE ASSESSMENT 4. Construct problems on electrostatics, electrodynamics and electromagnetism. Show solutions to such problems. 30 points** **( done by groups. Each group constructs problems on a certain topic ( draw lots )** **In preparation for their summative assessment 4** **==================================================================================================**