Chapter 26.pdf

# Electromotive force (EMF) and terminal voltage Electromotive force (EMF) represents the voltage produced by a source, such as a battery or generator, which is necessary to drive current through an electric circuit [5](#page=5). ### 1.1 Understanding electromotive force (EMF) EMF is defined as the energy per unit charge converted into electrical energy by the source. It is the ideal voltage provided by a source when no current is flowing. Sources like batteries or generators are essential for initiating and sustaining an electric current [5](#page=5). ### 1.2 Internal resistance and its effect on terminal voltage Real-world voltage sources, particularly batteries, possess a small internal resistance. This internal resistance acts as if it were connected in series with the ideal EMF of the source [5](#page=5) [6](#page=6). The presence of internal resistance leads to a reduction in the actual voltage available at the terminals of the source, known as the terminal voltage. When current flows through the internal resistance, a voltage drop occurs, diminishing the voltage that can be supplied to the external circuit [5](#page=5). The relationship between EMF, internal resistance, and terminal voltage can be expressed mathematically. If $\mathcal{E}$ represents the EMF, $r$ is the internal resistance, and $I$ is the current flowing through the source, then the terminal voltage $V_{terminal}$ is given by: $$V_{terminal} = \mathcal{E} - Ir$$ This formula indicates that the terminal voltage is equal to the EMF minus the voltage drop across the internal resistance [5](#page=5). #### 1.2.1 Calculating circuit parameters with internal resistance Consider a circuit where an external resistor $R$ is connected to a battery with EMF $\mathcal{E}$ and internal resistance $r$. The total resistance in the circuit is the sum of the external resistance and the internal resistance: $$R_{total} = R + r$$ The current $I$ flowing through the circuit can then be calculated using Ohm's Law: $$I = \frac{\mathcal{E}}{R_{total}} = \frac{\mathcal{E}}{R + r}$$ The terminal voltage $V_{ab}$ across the battery terminals can be determined in two ways: 1. Using the EMF and internal resistance: $$V_{ab} = \mathcal{E} - Ir$$ 2. Using the external resistor and the current: $$V_{ab} = IR$$ Both methods should yield the same result for the terminal voltage. #### 1.2.2 Power dissipation in a circuit with internal resistance The power dissipated in the external resistor $R$ is given by: $$P_R = I^2 R$$ The power dissipated within the battery's internal resistance $r$ is given by: $$P_r = I^2 r$$ The total power delivered by the EMF is the sum of the power dissipated in the external resistor and the internal resistance: $$P_{total} = P_R + P_r = I^2 R + I^2 r = I^2 (R+r)$$ This is also equal to $\mathcal{E}I$. > **Example:** A 65.0-ohm resistor is connected to the terminals of a battery with an EMF of 12.0 V and an internal resistance of 0.5 ohms. > (a) To find the current in the circuit, we use the formula $I = \frac{\mathcal{E}}{R + r}$. > $$I = \frac{12.0 \text{ V}}{65.0 \text{ \(\Omega\)} + 0.5 \text{ \(\Omega\)}} = \frac{12.0 \text{ V}}{65.5 \text{ \(\Omega\)}} \approx 0.183 \text{ A}$$ [7](#page=7). > (b) To find the terminal voltage, $V_{ab}$, we can use $V_{ab} = IR$. > $$V_{ab} = (0.183 \text{ A})(65.0 \text{ \(\Omega\)}) \approx 11.9 \text{ V}$$ [7](#page=7). > Alternatively, using $V_{ab} = \mathcal{E} - Ir$: > $$V_{ab} = 12.0 \text{ V} - (0.183 \text{ A})(0.5 \text{ \(\Omega\)}) \approx 12.0 \text{ V} - 0.0915 \text{ V} \approx 11.9 \text{ V}$$ [7](#page=7). > (c) The power dissipated in the resistor $R$ is $P_R = I^2 R$. > $$P_R = (0.183 \text{ A})^2 (65.0 \text{ \(\Omega\)}) \approx (0.0335 \text{ A}^2)(65.0 \text{ \(\Omega\)}) \approx 2.18 \text{ W}$$ [7](#page=7). > The power dissipated in the battery's internal resistance $r$ is $P_r = I^2 r$. > $$P_r = (0.183 \text{ A})^2 (0.5 \text{ \(\Omega\)}) \approx (0.0335 \text{ A}^2)(0.5 \text{ \(\Omega\)}) \approx 0.0168 \text{ W}$$ [7](#page=7). --- # Resistors in series and parallel configurations This topic explores the fundamental principles governing how resistors behave when connected in series and parallel configurations within an electrical circuit, focusing on calculating equivalent resistance and understanding circuit behavior [8](#page=8). ### 2.1 Resistors in series #### 2.1.1 Concept and characteristics In a series connection, circuit elements are arranged in a single path, so the current flows through each component sequentially before returning to the source. This means the current through each resistor is identical. However, the voltage across each resistor depends on its individual resistance value. The sum of the voltage drops across all resistors in series must equal the total voltage supplied by the battery [8](#page=8) [9](#page=9). #### 2.1.2 Calculating equivalent resistance in series The equivalent resistance ($R_{eq}$) for a series combination is the single resistance value that would yield the same total current in the circuit as the original configuration. It is calculated by simply summing the resistances of all individual resistors [10](#page=10): $$R_{eq} = R_1 + R_2 + R_3 + \dots$$ [10](#page=10). This implies that the total resistance in a series circuit is always greater than the largest individual resistance [10](#page=10). #### 2.1.3 Implications of series connection * **Current:** The current is the same through all components [9](#page=9). * **Voltage:** The total voltage is divided among the components, with larger resistances dropping more voltage [9](#page=9). * **Equivalent resistance:** $R_{eq}$ increases with each added resistor [10](#page=10). > **Tip:** When analyzing series circuits, remember that opening any part of the single path will break the entire circuit. ### 2.2 Resistors in parallel #### 2.2.1 Concept and characteristics In a parallel connection, the current splits, creating multiple paths for the electricity to flow through. Each component is connected across the same two points, ensuring that the voltage across each resistor is identical to the source voltage. The total current flowing out of the source is the sum of the currents passing through each individual resistor [11](#page=11) [12](#page=12). An analogy for parallel circuits involves water flow: the water (current) divides into separate streams (branches), each falling the same height (voltage). The total water flow is the sum of the individual stream flows. With multiple pipes open in parallel, the overall resistance to water flow is reduced compared to a single pipe [14](#page=14). #### 2.2.2 Calculating equivalent resistance in parallel For a parallel arrangement, the reciprocal of the equivalent resistance ($1/R_{eq}$) is equal to the sum of the reciprocals of the individual resistances [13](#page=13). $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$ [13](#page=13). For the special case of only two resistors in parallel, this simplifies to: $$R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2}$$ [13](#page=13). This formula highlights that the equivalent resistance in a parallel circuit is always less than the smallest individual resistance [13](#page=13). #### 2.2.3 Implications of parallel connection * **Current:** The total current is divided among the branches, with branches having lower resistance drawing more current [12](#page=12). * **Voltage:** The voltage is the same across all components [11](#page=11). * **Equivalent resistance:** $R_{eq}$ decreases as more resistors are added in parallel [13](#page=13). > **Tip:** In parallel circuits, if one component fails (e.g., a bulb burns out), the other components in parallel with it will generally continue to function because the current still has alternative paths. ### 2.3 Combined series and parallel circuits Many real-world circuits consist of combinations of resistors connected in both series and parallel configurations. Analyzing these circuits involves breaking them down into simpler series and parallel sections. This is typically done by first calculating the equivalent resistance of parallel groups, treating them as single resistors, and then combining these with any series resistors. This iterative process allows for the determination of total current, voltage drops, and current through individual components [17](#page=17) [18](#page=18). > **Example:** To find the total current drawn from a battery in a complex circuit, one would first simplify any parallel resistor combinations into their equivalent resistances, then combine these with series resistors to find the overall equivalent resistance of the entire circuit. Finally, Ohm's Law ($I = V/R_{eq}$) would be applied using the battery voltage and the total equivalent resistance to find the total current drawn [17](#page=17). ### 2.4 Applications and conceptual examples #### 2.4.1 Lightbulb brightness The brightness of a lightbulb is directly related to the power it dissipates, which in turn depends on the current flowing through it and the voltage across it ($P = I^2R$ or $P = V^2/R$). * **Series connection of identical bulbs:** In a series circuit with identical bulbs, the voltage drops equally across each bulb. Since the current is the same for all, the power dissipated by each is equal, meaning they will glow with equal brightness. If the resistances were different, the bulb with the higher resistance would drop more voltage and dissipate more power, thus glowing brighter [15](#page=15). * **Parallel connection of identical bulbs:** In a parallel circuit, each bulb receives the full source voltage. If the bulbs are identical, they will all dissipate the same amount of power and glow with the same brightness. A parallel configuration generally allows for more total light output than a series configuration for the same set of bulbs, as each bulb operates at full voltage [15](#page=15). * **Example with different wattages:** When a 100-W, 120-V bulb and a 60-W, 120-V bulb are connected in series, the 60-W bulb (which has a higher resistance) will glow more brightly because it dissipates more power due to the voltage division. In parallel, both bulbs would operate at their rated power and brightness, with the 100-W bulb being brighter [16](#page=16). #### 2.4.2 Multi-speed fans Resistors can be placed in series with a motor to control its speed. By adding resistance, the current drawn by the motor is reduced, causing it to run slower. For example, to reduce the current from 5.0 A to 2.0 A in a motor connected to a 12-V battery, a specific series resistor can be calculated using Ohm's Law and Kirchhoff's Voltage Law. The power rating required for this resistor must also be considered to prevent overheating [20](#page=20). #### 2.4.3 Circuit analysis with internal resistance When analyzing circuits, it's important to consider the internal resistance of the power source (e.g., a battery), denoted by $r$. This internal resistance reduces the terminal voltage of the battery available to the external circuit. The total current drawn from the battery is determined by the total resistance (external plus internal) and the terminal voltage is the source voltage minus the voltage drop across the internal resistance [21](#page=21). > **Example:** A 9.0-V battery with an internal resistance of 0.50 Ωconnected to an external circuit. To find the current drawn from the battery, one would first calculate the total equivalent resistance of the external circuit. Then, the total resistance of the circuit (external $R_{eq}$ + internal $r$) is used with the battery voltage to find the total current. The terminal voltage is calculated as $V_{terminal} = V_{battery} - I \cdot r$ [21](#page=21). --- # Kirchhoff's rules for circuit analysis Kirchhoff's rules provide a systematic method for analyzing complex electrical circuits that cannot be simplified into basic series and parallel combinations [22](#page=22). ### 3.1 The junction rule The junction rule, also known as Kirchhoff's current law (KCL), states that the sum of the currents entering a junction must equal the sum of the currents leaving that junction. This rule is a direct consequence of the conservation of electric charge. At any point where wires connect, charge cannot accumulate, so the rate at which charge flows in must equal the rate at which it flows out [23](#page=23). Mathematically, for a junction with currents $I_1, I_2, ..., I_n$: $$ \sum_{\text{entering}} I_i = \sum_{\text{leaving}} I_i $$ [23](#page=23). ### 3.2 The loop rule The loop rule, also known as Kirchhoff's voltage law (KVL), states that the algebraic sum of the changes in electric potential around any closed loop in a circuit must be zero. This principle is derived from the conservation of energy. As a charge moves around a closed path and returns to its starting point, the net work done on it must be zero, which implies that the net change in its potential energy (and thus potential) is zero [24](#page=24). When applying the loop rule, one must establish a direction for traversing the loop (e.g., clockwise or counterclockwise). The sign conventions for voltage changes across circuit elements are as follows: * **Resistors:** * If traversing in the same direction as the assumed current, the potential change is negative (voltage drop): $-\text{IR}$. * If traversing in the opposite direction of the assumed current, the potential change is positive (voltage rise): $+\text{IR}$. * **Batteries (EMFs):** * If traversing from the negative terminal to the positive terminal, the potential change is positive: $+\text{EMF}$. * If traversing from the positive terminal to the negative terminal, the potential change is negative: $-\text{EMF}$. The general form of the loop rule is: $$ \sum \Delta V = 0 $$ [24](#page=24). ### 3.3 Problem-solving strategy Applying Kirchhoff's rules effectively requires a systematic approach: 1. **Label currents and directions:** Assign a symbol (e.g., $I_1, I_2, I_3$) to each unknown current in the circuit. For each current, choose an assumed direction. It is crucial to be consistent with these chosen directions [25](#page=25). 2. **Identify unknowns:** Determine how many unknown currents or voltages exist in the circuit [25](#page=25). 3. **Apply rules to form equations:** * Apply the junction rule to as many junctions as needed to account for all currents. The number of independent junction rule equations will be one less than the number of junctions. * Apply the loop rule to independent closed loops in the circuit. The number of independent loop rule equations needed is equal to the total number of unknown currents minus the number of independent junction rule equations. In total, you will need as many independent equations as there are unknowns [25](#page=25). 4. **Solve the system of equations:** Solve the resulting system of linear equations for the unknown currents [25](#page=25). 5. **Interpret results:** If the calculated value for a current is negative, it means that the actual direction of the current is opposite to the direction you initially assumed [25](#page=25). > **Tip:** When drawing circuit diagrams for Kirchhoff's rules, it's helpful to clearly mark junctions and choose specific paths for traversing loops. Labeling voltage sources with their polarities is also essential. > **Example:** Consider a circuit with two loops and three unknown currents. You would typically apply the junction rule at one junction to get one equation. Then, you would apply the loop rule to each of the two independent loops to get two more equations, resulting in a total of three equations to solve for the three unknown currents. If a calculated current comes out as -2 amperes, and you assumed it flowed to the right, it actually flows to the left at 2 amperes [25](#page=25). --- # Series and parallel EMFs, and battery charging This section explores how multiple electromotive force (EMF) sources behave when connected in series and parallel configurations, including the nuanced scenario of one battery charging another. ### 4.1 EMFs in series When multiple EMF sources are connected in series, their voltages can either add up or subtract, depending on their polarities [27](#page=27). #### 4.1.1 EMFs in series in the same direction If several EMF sources are connected in series such that their EMFs act in the same direction, the total EMF of the combination is the sum of the individual EMFs [27](#page=27). > **Tip:** This is the most straightforward series combination where the total voltage is simply the sum of individual voltages. #### 4.1.2 EMFs in series in opposite directions When EMF sources are connected in series but in opposite directions, the total EMF is the difference between the individual EMFs. In such a configuration, the battery with the lower voltage will be charged by the battery with the higher voltage [28](#page=28). > **Example:** Consider two batteries, one with an EMF of 12 volts and another with an EMF of 9 volts, connected in series with opposite polarities. The net EMF will be $12 \text{ V} - 9 \text{ V} = 3 \text{ V}$. The 9-volt battery will be charged by the 12-volt battery. ### 4.2 EMFs in parallel Connecting EMF sources in parallel is typically meaningful only when all the sources have the same voltage. This arrangement allows for a greater total current to be supplied compared to what a single EMF source could provide [29](#page=29). > **Tip:** Parallel connections of EMFs with different voltages are generally not recommended due to potential damage or inefficient operation. ### 4.3 Battery charging scenarios Battery charging often involves connecting EMF sources in series, where one source (e.g., a good battery or a charger) provides power to another (e.g., a weak battery). The principle of EMFs in series in opposite directions is directly applicable here. #### 4.3.1 Jump starting a car A common practical example of battery charging is jump-starting a car. In this scenario, a good car battery (acting as a voltage source) is used to provide current to a weak battery and the car's starter motor [30](#page=30). The total resistance in the circuit when jump-starting includes the internal resistances of both batteries, the resistance of the jumper cables, and the resistance of the starter motor [30](#page=30). Let $\mathcal{E}_{\text{good}}$ and $r_{\text{good}}$ be the EMF and internal resistance of the good battery, respectively. Let $\mathcal{E}_{\text{weak}}$ and $r_{\text{weak}}$ be the EMF and internal resistance of the weak battery, respectively. Let $R_{\text{cables}}$ be the total resistance of the jumper cables, and $R_s$ be the resistance of the starter motor [30](#page=30). When the good battery is connected to the weak battery and the starter motor, the circuit behaves like two EMFs in series with opposite polarities. The current $I$ through the starter motor can be calculated using Ohm's Law for the entire circuit: $$I = \frac{\mathcal{E}_{\text{good}} - \mathcal{E}_{\text{weak}}}{\text{Total resistance of the circuit}}$$ The total resistance in this case is $r_{\text{good}} + r_{\text{weak}} + R_{\text{cables}} + R_s$. Therefore, the current is [30](#page=30): $$I = \frac{\mathcal{E}_{\text{good}} - \mathcal{E}_{\text{weak}}}{r_{\text{good}} + r_{\text{weak}} + R_{\text{cables}} + R_s}$$ > **Example 26-10 (from document):** A good car battery has an EMF of 12.5 V and an internal resistance of 0.020 Ω. A weak battery has an EMF of 10.1 V and an internal resistance of 0.10 Ω. Jumper cables have a combined resistance, and the starter motor has a resistance $R_s = 0.15$ Ω. To determine the current through the starter motor, the resistance of the jumper cables would need to be calculated or provided. Assuming the cables have some resistance, let's say $R_{\text{cables}} = 0.05$ Ω for illustration [30](#page=30). > (a) If only the weak battery is connected to the starter motor (assuming it could provide current), the current would be $I = \frac{\mathcal{E}_{\text{weak}}}{r_{\text{weak}} + R_s} = \frac{10.1 \text{ V}}{0.10 \Omega + 0.15 \Omega} = \frac{10.1}{0.25} \text{ A} = 40.4 \text{ A}$. However, this scenario doesn't involve battery charging. > (b) If the good battery is connected to jump-start the car with the weak battery and starter motor: > Total resistance = $r_{\text{good}} + r_{\text{weak}} + R_{\text{cables}} + R_s = 0.020 \Omega + 0.10 \Omega + 0.05 \Omega + 0.15 \Omega = 0.27 \Omega$. > Current through the starter motor = $I = \frac{12.5 \text{ V} - 10.1 \text{ V}}{0.27 \Omega} = \frac{2.4 \text{ V}}{0.27 \Omega} \approx 8.89 \text{ A}$. In this case, the weak battery is being charged while also supplying some current through its internal resistance. The higher EMF battery drives current through the lower EMF battery, effectively charging it. --- # RC circuits and their time constants This topic examines circuits containing resistors and capacitors, detailing the charging and discharging processes and the significance of the time constant. ### 5.1 Introduction to RC circuits RC circuits, consisting of resistors and capacitors, are fundamental in understanding transient electrical behavior. When a switch is closed in such a circuit, a capacitor begins to charge, leading to an increase in voltage across the capacitor and a corresponding decrease in current through the resistor [31](#page=31). ### 5.2 Charging a capacitor in an RC circuit In an RC circuit with an electromotive force (emf), the capacitor charges when a switch is closed. The voltage changes around the loop can be described by an equation. To find the charge as a function of time, this equation can be integrated [31](#page=31) [32](#page=32). The voltage across the capacitor ($V_C$) at any given time ($t$) is related to the charge ($Q$) by $V_C = \frac{Q}{C}$. The charge on the capacitor during charging can be expressed as a function of time using the following formula [33](#page=33): $$Q(t) = C\mathcal{E}(1 - e^{-t/RC})$$ where $C$ is the capacitance, $\mathcal{E}$ is the emf of the source, and $R$ is the resistance [33](#page=33). The quantity $RC$ that appears in the exponent is defined as the **time constant** of the circuit, often denoted by the Greek letter tau ($\tau$) [33](#page=33). $$ \tau = RC $$ The time constant has units of time and characterizes how quickly the capacitor charges or discharges [33](#page=33). The current ($I$) in the circuit at any time $t$ can be found by differentiating the charge with respect to time: $$I(t) = \frac{dQ}{dt} = \frac{\mathcal{E}}{R} e^{-t/RC}$$ This can also be expressed as: $$I(t) = I_{max} e^{-t/RC}$$ where $I_{max} = \frac{\mathcal{E}}{R}$ is the maximum initial current [34](#page=34). > **Tip:** After one time constant ($\tau$), the capacitor will be charged to approximately 63.2% of its maximum charge, and the current will have decreased to approximately 36.8% of its initial maximum value. After five time constants ($5\tau$), the capacitor is considered to be almost fully charged (about 99.3% of maximum charge), and the current is nearly zero. #### 5.2.1 Example: Charging RC circuit with emf **Example 26-11:** An RC circuit has a capacitance of $C = 0.30 \text{ } \mu\text{F}$, a total resistance of $R = 20 \text{ k}\Omega$, and a battery emf of $\mathcal{E} = 12 \text{ V}$ [35](#page=35). (a) Determine the time constant: $$ \tau = RC = (20 \times 10^3 \Omega) \times (0.30 \times 10^{-6} \text{ F}) = 6.0 \times 10^{-3} \text{ s} = 6.0 \text{ ms} $$ (b) Determine the maximum charge the capacitor could acquire: The maximum charge ($Q_{max}$) is given by $Q_{max} = C\mathcal{E}$. $$ Q_{max} = (0.30 \times 10^{-6} \text{ F}) \times (12 \text{ V}) = 3.6 \times 10^{-6} \text{ C} = 3.6 \text{ } \mu\text{C} $$ (c) Determine the time it takes for the charge to reach 99% of this value: We need to find $t$ when $Q(t) = 0.99 \times Q_{max}$. $$ 0.99 C\mathcal{E} = C\mathcal{E}(1 - e^{-t/\tau}) $$ $$ 0.99 = 1 - e^{-t/\tau} $$ $$ e^{-t/\tau} = 0.01 $$ $$ -t/\tau = \ln(0.01) $$ $$ t = -\tau \ln(0.01) = -(6.0 \text{ ms}) \times (-4.605) \approx 27.6 \text{ ms} $$ (d) Determine the current $I$ when the charge $Q$ is half its maximum value: When $Q = \frac{1}{2} Q_{max}$, we have $\frac{1}{2} C\mathcal{E} = C\mathcal{E}(1 - e^{-t/\tau})$, so $0.5 = 1 - e^{-t/\tau}$, which means $e^{-t/\tau} = 0.5$. The current is $I(t) = \frac{\mathcal{E}}{R} e^{-t/\tau}$. $$ I = \frac{12 \text{ V}}{20 \times 10^3 \Omega} \times 0.5 = (0.60 \times 10^{-3} \text{ A}) \times 0.5 = 0.30 \text{ mA} $$ (e) Determine the maximum current: The maximum current occurs at $t=0$ and is given by $I_{max} = \frac{\mathcal{E}}{R}$. $$ I_{max} = \frac{12 \text{ V}}{20 \times 10^3 \Omega} = 0.60 \times 10^{-3} \text{ A} = 0.60 \text{ mA} $$ (f) Determine the charge $Q$ when the current $I$ is 0.20 of its maximum value: If $I = 0.20 I_{max}$, then from $I(t) = I_{max} e^{-t/RC}$, we have $0.20 = e^{-t/RC}$, so $e^{-t/RC} = 0.20$. The charge is $Q(t) = C\mathcal{E}(1 - e^{-t/RC})$. $$ Q = C\mathcal{E}(1 - 0.20) = C\mathcal{E}(0.80) = 0.80 Q_{max} $$ $$ Q = 0.80 \times (3.6 \text{ } \mu\text{C}) = 2.88 \text{ } \mu\text{C} $$ ### 5.3 Discharging a capacitor in an RC circuit When a charged capacitor is disconnected from a power source and connected across a resistor, it begins to discharge. The voltage across the capacitor decreases exponentially with time, and the current through the resistor also decreases exponentially [36](#page=36) [37](#page=37). The charge on the capacitor during discharge is given by: $$Q(t) = Q_0 e^{-t/RC}$$ where $Q_0$ is the initial charge on the capacitor at $t=0$ [37](#page=37). The voltage across the capacitor during discharge is: $$V_C(t) = \frac{Q_0}{C} e^{-t/RC} = V_0 e^{-t/RC}$$ where $V_0 = \frac{Q_0}{C}$ is the initial voltage across the capacitor [37](#page=37). The current through the resistor during discharge is: $$I(t) = -\frac{dQ}{dt} = -\frac{Q_0}{RC} e^{-t/RC} = -\frac{V_0}{R} e^{-t/RC}$$ The negative sign indicates that the current flows in a direction that reduces the charge on the capacitor. The magnitude of the current is [37](#page=37): $$|I(t)| = \frac{V_0}{R} e^{-t/RC} = |I_{max}| e^{-t/RC}$$ where $|I_{max}| = \frac{V_0}{R}$ is the initial magnitude of the current. > **Tip:** The time constant $\tau = RC$ also governs the rate of discharge. After one time constant, the charge and voltage will have decreased to approximately 36.8% of their initial values, and the current magnitude will have similarly decreased. #### 5.3.1 Example: Discharging RC circuit **Example 26-12:** In an RC circuit, a battery has fully charged a capacitor. At $t=0$, a switch is thrown from position 'a' (connecting to the battery) to position 'b' (connecting to a resistor). The battery emf is $\mathcal{E} = 20.0 \text{ V}$, and the capacitance is $C = 1.02 \text{ } \mu\text{F}$. The current $I$ is observed to decrease to 0.50 of its initial value in $40 \text{ } \mu\text{s}$ [38](#page=38). (a) What is the value of $Q_0$, the charge on the capacitor, at $t=0$? At $t=0$, the capacitor is fully charged by the battery. $$ Q_0 = C\mathcal{E} = (1.02 \times 10^{-6} \text{ F}) \times (20.0 \text{ V}) = 20.4 \times 10^{-6} \text{ C} = 20.4 \text{ } \mu\text{C} $$ (b) What is the value of $R$? The current during discharge is given by $|I(t)| = \frac{V_0}{R} e^{-t/RC}$, where $V_0 = \mathcal{E} = 20.0 \text{ V}$ (since the capacitor was fully charged). We are given that $|I(t)| = 0.50 |I_{max}|$ at $t = 40 \text{ } \mu\text{s}$. $$ 0.50 |I_{max}| = |I_{max}| e^{-(40 \times 10^{-6} \text{ s}) / (R \times 1.02 \times 10^{-6} \text{ F})} $$ $$ 0.50 = e^{-(40 \times 10^{-6}) / (R \times 1.02 \times 10^{-6})} $$ Taking the natural logarithm of both sides: $$ \ln(0.50) = -\frac{40 \times 10^{-6}}{R \times 1.02 \times 10^{-6}} $$ $$ -0.6931 = -\frac{40}{R \times 1.02} $$ $$ R = \frac{40}{0.6931 \times 1.02} \approx \frac{40}{0.70696} \approx 56.58 \Omega $$ (c) What is $Q$ at $t=60 \text{ } \mu\text{s}$? First, calculate the time constant: $$ \tau = RC = (56.58 \Omega) \times (1.02 \times 10^{-6} \text{ F}) \approx 57.71 \times 10^{-6} \text{ s} = 57.71 \text{ } \mu\text{s} $$ Now, calculate the charge at $t=60 \text{ } \mu\text{s}$: $$ Q(60 \text{ } \mu\text{s}) = Q_0 e^{-(60 \text{ } \mu\text{s}) / (57.71 \text{ } \mu\text{s})} $$ $$ Q(60 \text{ } \mu\text{s}) = (20.4 \text{ } \mu\text{C}) e^{-1.040} $$ $$ Q(60 \text{ } \mu\text{s}) \approx (20.4 \text{ } \mu\text{C}) \times 0.3532 \approx 7.21 \text{ } \mu\text{C} $$ ### 5.4 Applications and Conceptual Examples #### 5.4.1 Bulb in an RC circuit **Conceptual Example 26-13:** Consider an RC circuit with a capacitor that is initially uncharged, connected to a resistor and a lightbulb in series with a switch. When the switch is closed, the capacitor begins to charge. Initially, the current is high, causing the bulb to glow brightly. As the capacitor charges, the current decreases exponentially. Consequently, the brightness of the lightbulb will fade over time until it is no longer illuminated if the capacitor is fully charged and there is no leakage. If the capacitor is then discharged, the bulb will briefly glow again as current flows out of the capacitor [39](#page=39). #### 5.4.2 Resistor in a turn signal **Example 26-14:** The order of magnitude of a resistor in a turn-signal circuit can be estimated. Turn signals typically flash at a rate of about once or twice per second. This flashing is often achieved using an RC circuit where the charging and discharging times are on the order of a second. This implies that the time constant ($\tau = RC$) is roughly in the range of fractions of a second to a few seconds [40](#page=40). --- # Electrical hazards and safety precautions Electrical hazards pose significant dangers, primarily through electric shock and burns, emphasizing the critical need for stringent safety measures when dealing with electric currents and voltages [41](#page=41). ### 6.1 Understanding electrical hazards #### 6.1.1 Effects of electric current on the human body The severity of an electric shock is directly related to the amount of current flowing through the body and the duration of contact. * **Perceptible current:** Even a small current of approximately 1 milliampere (mA) can be felt by most individuals [41](#page=41). * **Painful currents:** Currents in the range of a few milliamperes can cause pain [41](#page=41). * **Uncontrollable muscle contractions:** Currents exceeding 10 mA can lead to muscle spasms so strong that a person may be unable to release their grip, making self-rescue difficult [41](#page=41). * **Ventricular fibrillation:** A current of around 100 mA passing through the torso can disrupt the heart's rhythm, causing ventricular fibrillation and potentially leading to death [41](#page=41). * **Burns:** Higher current levels, even if they don't cause fibrillation, can result in severe burns [41](#page=41). > **Tip:** It's crucial to remember that a person receiving an electric shock has inadvertently become part of a complete electrical circuit, allowing current to flow through them [42](#page=42). #### 6.1.2 Factors increasing electrical risk Certain conditions can significantly increase the danger posed by household voltages. * **Wet conditions:** Household voltages can be lethal when a person is wet and has good contact with the ground. This is because water reduces the body's electrical resistance [41](#page=41). * **Faulty wiring and improper grounding:** Defective electrical connections and inadequate grounding systems create pathways for current to flow unexpectedly, posing a serious risk [43](#page=43). > **Tip:** Always be cautious when working with electricity, especially in environments where moisture is present or where electrical systems may be compromised [41](#page=41) [43](#page=43). ### 6.2 Safety precautions and best practices Implementing proper safety measures is paramount to preventing electrical accidents. #### 6.2.1 Professional electrical work * Ensuring electrical work is performed by qualified professionals is a key safety measure. They possess the knowledge and skills to identify hazards and implement appropriate safety standards [43](#page=43). #### 6.2.2 Understanding wiring and grounding * **Three-prong plugs:** The safest types of electrical plugs are those with three prongs. The third prong provides a separate ground line, which offers an additional layer of protection by diverting fault current safely to the ground [44](#page=44). * **Identifying the hot wire:** Before performing any electrical tasks, it is essential to correctly identify the "hot" wire. This wire carries the voltage and is the primary source of electrical danger. Electrical wiring colors can vary, so always confirm with proper testing equipment or consult a professional [44](#page=44). > **Example:** In some wiring configurations, the hot wire might be black or red, the neutral wire white, and the ground wire green or bare copper. However, these colors are not universal, and relying solely on them can be dangerous. Always verify with a multimeter or by consulting an electrician [44](#page=44). --- # Ammeters, voltmeters, and ohmmeters This topic explores the function, design principles, and proper circuit connection of ammeters, voltmeters, and ohmmeters, which are essential electrical measurement instruments. ### 7.1 Ammeters An ammeter is a device used to measure the electric current flowing through a circuit. To accurately measure current without significantly altering it, an ammeter must be connected in series with the circuit element through which the current is to be measured. Consequently, ammeters are designed to have very low internal resistance. This low resistance minimizes the voltage drop across the ammeter itself, ensuring that the current being measured is not substantially affected [45](#page=45) [50](#page=50). #### 7.1.1 Ammeter design principles The design of an ammeter often utilizes a galvanometer, which is sensitive to small currents. A galvanometer typically has a coil with a certain internal resistance, denoted as $r$. To convert a galvanometer into an ammeter with a specific full-scale current range, a low-resistance resistor, known as a shunt resistor ($R_{\text{shunt}}$), is connected in parallel with the galvanometer. This shunt resistor diverts most of the current around the galvanometer coil, allowing the ammeter to measure larger currents [45](#page=45) [46](#page=46). The relationship between the galvanometer's full-scale current ($I_{\text{fs}}$), its internal resistance ($r$), the desired full-scale current for the ammeter ($I_{\text{ammeter}}$), and the shunt resistance ($R_{\text{shunt}}$) can be derived. For the current to be at full scale ($I_{\text{ammeter}}$), the current through the galvanometer will be its full-scale sensitivity ($I_{\text{fs}}$), and the remaining current will flow through the shunt resistor. Since the shunt resistor and the galvanometer are in parallel, the voltage drop across them must be equal: $I_{\text{fs}} \cdot r = (I_{\text{ammeter}} - I_{\text{fs}}) \cdot R_{\text{shunt}}$ From this equation, the required shunt resistance can be calculated: $$ R_{\text{shunt}} = \frac{I_{\text{fs}} \cdot r}{I_{\text{ammeter}} - I_{\text{fs}}} $$ The scale of an ammeter designed this way is linear if the galvanometer's deflection is directly proportional to the current passing through it [46](#page=46) [48](#page=48). #### 7.1.2 Example: Ammeter design Designing an ammeter to read 1.0 A at full scale using a galvanometer with a full-scale sensitivity of 50 μA (which is $50 \times 10^{-6}$ A) and an internal resistance $r = 30 \, \Omega$. The scale is linear as galvanometers typically provide linear scales with respect to current. The shunt resistance needed is calculated as: $$ R_{\text{shunt}} = \frac{(50 \times 10^{-6} \, \text{A}) \cdot (30 \, \Omega)}{(1.0 \, \text{A}) - (50 \times 10^{-6} \, \text{A})} \approx \frac{1.5 \times 10^{-3} \, \Omega}{1.0 \, \text{A}} = 1.5 \times 10^{-3} \, \Omega $$ So, a shunt resistor of approximately 0.0015 $\Omega$ would be needed [46](#page=46). ### 7.2 Voltmeters A voltmeter is an instrument used to measure the electric potential difference, or voltage, across two points in a circuit. To ensure that a voltmeter does not significantly alter the voltage it is intended to measure, it must be connected in parallel with the circuit element across which the voltage is being measured. Consequently, voltmeters are designed to have very high internal resistance. This high resistance ensures that only a negligible amount of current is drawn from the circuit, thereby minimally affecting the voltage distribution [45](#page=45) [47](#page=47) [50](#page=50). #### 7.2.1 Voltmeter design principles Similar to ammeters, voltmeters are often built using galvanometers. To convert a galvanometer into a voltmeter, a high-resistance resistor, known as a multiplier resistor ($R_{\text{multiplier}}$), is connected in series with the galvanometer coil. This series resistor increases the total resistance of the voltmeter, limiting the current that flows through the galvanometer when a voltage is applied across the voltmeter [45](#page=45) [48](#page=48). Let $I_{\text{fs}}$ be the full-scale current sensitivity of the galvanometer and $r$ be its internal resistance. If the voltmeter is designed to measure a maximum voltage of $V_{\text{voltmeter}}$ at full scale, then the total resistance of the voltmeter ($R_{\text{total}}$) is given by Ohm's Law: $R_{\text{total}} = \frac{V_{\text{voltmeter}}}{I_{\text{fs}}}$ The total resistance is the sum of the galvanometer's internal resistance and the multiplier resistor: $R_{\text{total}} = r + R_{\text{multiplier}}$ Therefore, the multiplier resistance can be calculated as: $$ R_{\text{multiplier}} = R_{\text{total}} - r = \frac{V_{\text{voltmeter}}}{I_{\text{fs}}} - r $$ The scale of a voltmeter designed this way is linear if the galvanometer's deflection is proportional to the current, which is directly proportional to the voltage across the voltmeter [48](#page=48). #### 7.2.2 Example: Voltmeter design Designing a voltmeter that reads from 0 to 15 V using a galvanometer with an internal resistance of $30 \, \Omega$ and a full-scale current sensitivity of 50 μA ($50 \times 10^{-6}$ A). The scale is linear. First, calculate the total resistance required for the 15 V full-scale reading: $$ R_{\text{total}} = \frac{15 \, \text{V}}{50 \times 10^{-6} \, \text{A}} = \frac{15}{5 \times 10^{-5}} \, \Omega = 3 \times 10^5 \, \Omega = 300,000 \, \Omega $$ Now, calculate the multiplier resistance: $$ R_{\text{multiplier}} = R_{\text{total}} - r = 300,000 \, \Omega - 30 \, \Omega = 299,970 \, \Omega $$ So, a multiplier resistor of approximately 299,970 $\Omega$ would be connected in series with the galvanometer [48](#page=48). #### 7.2.3 Voltmeter loading effect The finite, albeit large, resistance of a voltmeter can still affect the circuit it is measuring, especially in high-impedance circuits. This phenomenon is known as voltmeter loading. For example, if a voltmeter with a sensitivity of 10,000 $\Omega$/V is used on the 5.0-V scale, its total resistance is $5.0 \, \text{V} \times 10,000 \, \Omega/\text{V} = 50,000 \, \Omega$. If this voltmeter is connected across one of two identical 15 k$\Omega$ resistors in series with an 8.0 V battery, the true voltage across the resistor would be half the battery voltage (4.0 V) if the voltmeter had infinite resistance. However, with the 50,000 $\Omega$ voltmeter in parallel with the 15,000 $\Omega$ resistor, the equivalent resistance becomes: $$ R_{\text{eq}} = \frac{50,000 \, \Omega \times 15,000 \, \Omega}{50,000 \, \Omega + 15,000 \, \Omega} = \frac{750 \times 10^6 \, \Omega^2}{65,000 \, \Omega} \approx 11,538 \, \Omega $$ The total resistance of the circuit is now $11,538 \, \Omega + 15,000 \, \Omega = 26,538 \, \Omega$. The new total current is $8.0 \, \text{V} / 26,538 \, \Omega \approx 0.000301 \, \text{A}$ or 301 $\mu$A. The voltage read by the meter across the first resistor (which is now in parallel with the voltmeter) would be $I_{\text{total}} \cdot R_{\text{eq}} \approx 0.000301 \, \text{A} \times 11,538 \, \Omega \approx 3.47 \, \text{V}$. The error caused by the voltmeter's finite resistance is approximately $4.0 \, \text{V} - 3.47 \, \text{V} = 0.53 \, \text{V}$, or about a $13\%$ error [51](#page=51). > **Tip:** To minimize voltmeter loading error, use a voltmeter with a much higher resistance than the resistance of the circuit element across which you are measuring the voltage. ### 7.3 Ohmmeters An ohmmeter is an instrument used to measure electrical resistance. Unlike ammeters and voltmeters that measure quantities in an active circuit, ohmmeters require their own internal power source, typically a battery, to drive a current through the resistance being measured. This allows the ohmmeter to determine the resistance based on Ohm's Law ($R = V/I$) [49](#page=49). #### 7.3.1 Ohmmeter operation An ohmmeter typically consists of a galvanometer, a battery, and a multiplier resistor, similar to a voltmeter. However, the unknown resistance is connected externally to the ohmmeter terminals. The internal battery drives a current through the series combination of the galvanometer, multiplier resistor, and the unknown external resistance. The deflection of the galvanometer, and thus the reading on the scale, is inversely proportional to the unknown resistance. > **Tip:** Always ensure that the circuit being tested is de-energized and disconnected from any power source before connecting an ohmmeter to measure resistance. This prevents damage to the ohmmeter and ensures accurate readings. Digital ohmmeters are also common and employ more sophisticated electronic circuits for resistance measurement, often using the principle of measuring the voltage drop across a known internal resistor when the unknown resistance is placed in series with it, or by measuring the current flowing through the unknown resistance when a known voltage is applied. --- ## Common mistakes to avoid - Review all topics thoroughly before exams - Pay attention to formulas and key definitions - Practice with examples provided in each section - Don't memorize without understanding the underlying concepts

| Term | Definition | |------|------------| | Electromotive force (EMF) | The energy per unit charge converted from some other form to electrical energy by a battery or other source, typically measured in volts. It represents the ideal voltage of a source when no current is flowing. | | Terminal voltage | The actual voltage measured across the terminals of a source of emf when current is flowing. It is less than the emf due to the voltage drop across the internal resistance of the source. | | Internal resistance | The resistance inherent within a source of emf, such as a battery. This resistance causes a voltage drop when current flows, reducing the terminal voltage. | | Resistors in series | Components connected end-to-end, so that the same current flows through each component. The total resistance is the sum of individual resistances. | | Resistors in parallel | Components connected across the same two points, so that the voltage across each component is the same. The total current is the sum of the currents through each component. | | Equivalent resistance | A single resistance value that would produce the same total current and voltage drop as a combination of resistors in a circuit. | | Kirchhoff's junction rule | A statement that the algebraic sum of currents entering any junction (or node) in a circuit must equal the algebraic sum of currents leaving that junction. This rule is based on the conservation of electric charge. | | Kirchhoff's loop rule | A statement that the algebraic sum of the potential differences (voltages) around any closed loop in a circuit must be zero. This rule is based on the conservation of energy. | | Time constant (RC circuit) | A characteristic time for an RC circuit, denoted by the Greek letter tau ($$\tau$$), equal to the product of the resistance R and the capacitance C ($$\tau = RC$$). It represents the time it takes for the charge or current to decay to about 37% of its initial value during discharge, or to reach about 63% of its final value during charging. | | Ammeter | An instrument used to measure electric current. It must be connected in series with the circuit element through which the current is to be measured, and ideally has very low resistance. | | Voltmeter | An instrument used to measure electric potential difference (voltage). It must be connected in parallel with the circuit element across which the voltage is to be measured, and ideally has very high resistance. | | Ohmmeter | An instrument used to measure electrical resistance. It typically contains a battery to supply a small current and a meter to read the resulting resistance. | | Fibrillation | A serious medical condition where the heart muscle quivers instead of pulsing normally, often caused by electric currents passing through the torso. |