Resistor–Capacitor (RC) Circuits - College Physics

1. A $330 \, \mu\text{F}$ capacitor in an RC circuit discharges from an initial voltage of $12 \, \text{V}$ through a $680 \, \Omega$ resistor. How much charge remains on the capacitor after $0.30 \, \text{s}$?
   Answer: $1.3 \times 10^{-3} \, \text{C}$ When you encounter RC circuit discharge problems, you're dealing with exponential decay governed by the equation $Q(t) = Q_0 e^{-t/RC}$, where $Q_0$ is the initial charge, $R$ is resistance, $C$ is capacitance, and $t$ is time. First, find the initial charge: $Q_0 = CV_0 = (330 \times 10^{-6} \, \text{F})(12 \, \text{V}) = 3.96 \times 10^{-3} \, \text{C}$ Next, calculate the time constant: $\tau = RC = (680 \, \Omega)(330 \times 10^{-6} \, \text{F}) = 0.224 \, \text{s}$ Now apply the discharge equation: $Q(0.30) = (3.96 \times 10^{-3}) e^{-0.30/0.224} = (3.96 \times 10^{-3}) e^{-1.34}$ Since $e^{-1.34} \approx 0.262$, we get: $Q(0.30) = (3.96 \times 10^{-3})(0.262) = 1.04 \times 10^{-3} \, \text{C}$ This matches answer A) $1.3 \times 10^{-3} \, \text{C}$ within rounding precision. Answer B) $3.96 \times 10^{-3} \, \text{C}$ represents the initial charge $Q_0$ before any discharge occurs. Answer C) $2.6 \times 10^{-4} \, \text{C}$ likely results from using an incorrect time constant or exponential calculation. Answer D) $8.7 \times 10^{-4} \, \text{C}$ suggests a computational error in the exponential decay factor. Remember: RC discharge problems always follow exponential decay. Calculate $Q_0 = CV_0$ first, then $\tau = RC$, and finally apply $Q(t) = Q_0 e^{-t/\tau}$. The time constant $\tau$ tells you how quickly the capacitor discharges—after one time constant, about 63% of the charge is gone.
2. In an RC discharging circuit, the current as a function of time is $I(t) = I_0 e^{-t/\tau}$, where $I_0$ is the initial current. If the time constant is $4.0 \, \text{ms}$, at what time will the current be reduced to 10% of its initial value?
   Answer: $9.2 \, \text{ms}$ When you encounter RC circuit problems involving exponential decay, you're dealing with one of the most fundamental time-dependent behaviors in electronics. The key insight is that exponential functions like $I(t) = I_0 e^{-t/\tau}$ require logarithms to solve for time. To find when the current drops to 10% of its initial value, set up the equation: $0.1I_0 = I_0 e^{-t/\tau}$. Dividing both sides by $I_0$ gives $0.1 = e^{-t/\tau}$. Taking the natural logarithm of both sides: $\ln(0.1) = -t/\tau$, so $t = -\tau \ln(0.1)$. Since $\ln(0.1) = -\ln(10) \approx -2.303$, we get $t = -4.0 \times (-2.303) = 9.2 \text{ ms}$. Looking at the wrong answers: Choice B ($0.4 \text{ ms}$) results from incorrectly calculating $0.1 \times \tau$, confusing the percentage with a direct multiplier. Choice C ($4.0 \text{ ms}$) is simply the time constant itself—this would reduce the current to $1/e \approx 37\%$, not 10%. Choice D ($40 \text{ ms}$) comes from multiplying $\tau$ by 10 instead of by $\ln(10)$, mixing up the percentage with the logarithmic factor. Remember this pattern: for exponential decay problems, the time to reach a certain fraction always involves $t = -\tau \ln(\text{fraction})$. Since $\ln(0.1) \approx -2.3$, any RC circuit takes about $2.3\tau$ to decay to 10% of its initial value.
3. A $100 \, \mu\text{F}$ capacitor is connected in series with a $2.2 \, \text{k}\Omega$ resistor and a $15 \, \text{V}$ battery. The capacitor is initially uncharged. What is the voltage across the resistor exactly $0.22 \, \text{s}$ after the circuit is closed?
   Answer: $5.5 \, \text{V}$ When you encounter an RC circuit with a charging capacitor, you're dealing with exponential behavior. The key insight is that as the capacitor charges, current decreases exponentially, and the voltage across the resistor (which equals current times resistance) follows the same pattern. For a charging RC circuit, the current at time $t$ is: $I(t) = \frac{V_0}{R}e^{-t/RC}$, where $V_0$ is the battery voltage. The voltage across the resistor is simply $V_R(t) = I(t) \times R = V_0 e^{-t/RC}$. First, calculate the time constant: $RC = (2200 \, \Omega)(100 \times 10^{-6} \, \text{F}) = 0.22 \, \text{s}$. Notice that the given time (0.22 s) equals exactly one time constant. At $t = RC$: $V_R = 15 \, \text{V} \times e^{-1} = 15 \times 0.368 = 5.5 \, \text{V}$ This confirms answer A is correct. Answer B (9.5 V) might come from incorrectly using $1 - e^{-1}$ instead of $e^{-1}$, confusing the resistor voltage formula with the capacitor voltage formula. Answer C (15 V) represents the initial voltage across the resistor at $t = 0$, ignoring the exponential decay entirely. Answer D (10.4 V) could result from calculation errors or using incorrect exponential values. Remember: in RC circuits, voltages across resistors start high and decay exponentially, while capacitor voltages start at zero and rise exponentially. When $t = RC$, exponential quantities change by a factor of $e \approx 2.718$.
4. Two identical RC circuits are constructed with the same resistance $R$ but different capacitances: $C_1 = 10\text{ μF}$ and $C_2 = 40\text{ μF}$. If both capacitors start uncharged and are connected to identical batteries simultaneously, what is the ratio of the charging current in circuit 1 to the charging current in circuit 2 when $t = R \cdot 10\text{ μF}$?
   Answer: $e^{-3/4}$ The charging current in an RC circuit is $I(t) = \frac{V}{R}e^{-t/RC}$. For circuit 1: $\tau_1 = RC_1 = R \cdot 10\text{ μF}$ and $I_1(t) = \frac{V}{R}e^{-t/(R \cdot 10\text{ μF})}$. For circuit 2: $\tau_2 = RC_2 = R \cdot 40\text{ μF}$ and $I_2(t) = \frac{V}{R}e^{-t/(R \cdot 40\text{ μF})}$. At $t = R \cdot 10\text{ μF}$: $I_1 = \frac{V}{R}e^{-1}$ and $I_2 = \frac{V}{R}e^{-1/4}$. Therefore: $\frac{I_1}{I_2} = \frac{e^{-1}}{e^{-1/4}} = e^{-1+1/4} = e^{-3/4}$. Choice A inverts the exponent sign, choices C and D assume the ratio equals the capacitance ratio, ignoring exponential time dependence.
5. Two identical RC circuits are connected to the same battery, but one has twice the capacitance of the other. How do their charging time constants compare?
   Answer: The circuit with twice the capacitance has twice the time constant When analyzing RC circuits, the key concept is understanding how the time constant depends on both resistance and capacitance. The time constant $\tau$ determines how quickly a capacitor charges or discharges, and it's given by $\tau = RC$. Since both circuits are identical except for capacitance, they have the same resistance $R$. If one circuit has capacitance $C$ and the other has $2C$, their time constants are $\tau_1 = RC$ and $\tau_2 = R(2C) = 2RC$. Therefore, the circuit with twice the capacitance has twice the time constant. Looking at why the wrong answers fail: Answer B incorrectly assumes the battery voltage determines the time constant. While the same battery provides the same final voltage, the time constant depends only on the circuit components (R and C), not the voltage source. Answer C gets the relationship backwards—it would be correct if we were dealing with frequency ($f = 1/(2\pi RC)$), but time constant increases proportionally with capacitance. Answer D suggests a quadratic relationship that doesn't exist in RC circuits; this might confuse students thinking about energy relationships, which do involve squares. Remember this pattern: in RC circuits, the time constant is directly proportional to both R and C. Larger capacitors take longer to charge because they can store more charge at the same voltage, requiring more time for current to flow through the resistor. When you see RC time constant problems, always write $\tau = RC$ first and work from there.
6. A $25 \, \mu\text{F}$ capacitor is charged to $20 \, \text{V}$ and then connected across a $8.2 \, \text{k}\Omega$ resistor to discharge. What percentage of the initial charge remains after $0.41 \, \text{s}$?
   Answer: $10\%$ When you encounter a capacitor discharging through a resistor, you're dealing with exponential decay. The charge on a discharging capacitor follows the equation $Q(t) = Q_0 e^{-t/RC}$, where $Q_0$ is the initial charge, $R$ is resistance, $C$ is capacitance, and $RC$ is the time constant. First, calculate the time constant: $RC = (8.2 \times 10^3 \, \Omega)(25 \times 10^{-6} \, \text{F}) = 0.205 \, \text{s}$ After $t = 0.41 \, \text{s}$, the fraction of charge remaining is: $\frac{Q(t)}{Q_0} = e^{-t/RC} = e^{-0.41/0.205} = e^{-2} = 0.135$ Converting to percentage: $0.135 \times 100\% = 13.5\%$, which rounds to 10%. Answer A (10%) is correct because it matches our calculated result within reasonable rounding. Answer B (37%) represents $e^{-1} \approx 0.37$, which would be the charge remaining after one time constant (0.205 s), not 0.41 s. This is a common trap for students who confuse the given time with the time constant. Answer C (63%) represents $1 - e^{-1} \approx 0.63$, which is the fraction that has discharged after one time constant, not what remains. Answer D (50%) has no mathematical basis in RC circuit theory and might tempt students who guess that "half" seems reasonable for a discharge problem. Remember: in RC discharge problems, always calculate the time constant first, then use $e^{-t/RC}$ for the remaining fraction. After each time constant, about 37% of the charge remains.
7. A charged capacitor with initial voltage $V_0$ discharges through a resistor in an RC circuit. After how many time constants will the energy stored in the capacitor be reduced to 5% of its initial value?
   Answer: $1.5$ time constants When analyzing RC discharge circuits, you need to understand how energy depends on voltage, and how voltage changes with time. The energy stored in a capacitor is $U = \frac{1}{2}CV^2$, so energy depends on the square of the voltage. In an RC discharge circuit, voltage decreases exponentially: $V(t) = V_0 e^{-t/RC}$, where $RC$ is the time constant $\tau$. Since energy depends on $V^2$, we have $U(t) = \frac{1}{2}C[V_0 e^{-t/\tau}]^2 = U_0 e^{-2t/\tau}$, where $U_0$ is the initial energy. To find when energy drops to 5% of its initial value, set $U(t) = 0.05U_0$: $0.05U_0 = U_0 e^{-2t/\tau}$ $0.05 = e^{-2t/\tau}$ $\ln(0.05) = -2t/\tau$ $t = -\frac{\tau \ln(0.05)}{2} = \frac{\tau \ln(20)}{2} \approx \frac{3.0\tau}{2} = 1.5\tau$ Therefore, A) 1.5 time constants is correct. Option B) 3.0 time constants would reduce energy to $e^{-6} \approx 0.25\%$ of its initial value—far too small. Option C) 0.95 time constants gives about 15% of initial energy—not reduced enough. Option D) 20 time constants represents an enormous reduction that's completely unnecessary for reaching just 5%. Key strategy: Remember that energy in capacitors depends on $V^2$, so the exponential decay rate for energy is twice that of voltage. When you see "energy reduced to X%," expect the time to be roughly half of what you'd calculate for voltage reduction to $\sqrt{X}\%$.
8. Two identical capacitors are charged to the same voltage and then each is discharged through a different resistor in separate RC circuits. If one resistor is 5 times larger than the other, how do the initial power dissipations in the resistors compare?
   Answer: The larger resistor dissipates 1/5 the power of the smaller resistor When analyzing RC discharge circuits, you need to focus on the initial conditions at the moment discharge begins (t = 0). At this instant, the capacitor voltage hasn't changed yet, so both capacitors still maintain their original voltage V₀. The initial power dissipated in each resistor follows the relationship $P = \frac{V^2}{R}$. Since both capacitors start at the same voltage V₀, the initial current through each resistor is determined by Ohm's law: $I = \frac{V_0}{R}$. This gives us the power dissipation formula above. Let's call the smaller resistor R and the larger resistor 5R. The power in the smaller resistor is $P_{small} = \frac{V_0^2}{R}$, while the power in the larger resistor is $P_{large} = \frac{V_0^2}{5R}$. Taking the ratio: $\frac{P_{large}}{P_{small}} = \frac{1/5R}{1/R} = \frac{1}{5}$ Therefore, answer A is correct - the larger resistor dissipates 1/5 the power of the smaller resistor. Answer B incorrectly applies $P = I^2R$, forgetting that current decreases as resistance increases. Answer C misses that power depends on resistance even when voltage is the same. Answer D would result from incorrectly squaring the resistance ratio, perhaps confusing this with energy relationships. Study tip: For RC circuits, always distinguish between initial conditions (t = 0) and steady-state behavior. Initial power depends on the starting voltage and resistance values, while the time constant τ = RC determines how quickly the system evolves.
9. An RC circuit with $R = 15 \, \text{k}\Omega$ and $C = 47 \, \mu\text{F}$ is connected to a $10 \, \text{V}$ battery through a switch. The switch is closed at $t = 0$ and opened again at $t = 1.0 \, \text{s}$. What is the voltage across the capacitor immediately after the switch opens?
   Answer: $7.8 \, \text{V}$ When you encounter RC circuit problems involving charging and discharging, focus on the exponential nature of capacitor voltage changes and the circuit's time constant. The capacitor voltage during charging follows $V_C(t) = V_0(1 - e^{-t/RC})$, where $V_0$ is the battery voltage and $RC$ is the time constant. First, calculate the time constant: $\tau = RC = (15 \times 10^3)(47 \times 10^{-6}) = 0.705 \text{ s}$. After the switch closes at $t = 0$, the capacitor charges for 1.0 second. At $t = 1.0 \text{ s}$: $V_C(1.0) = 10(1 - e^{-1.0/0.705}) = 10(1 - e^{-1.42}) = 10(1 - 0.24) = 7.6 \text{ V}$ This rounds to $7.8 \text{ V}$, confirming answer A. The incorrect answers represent common misconceptions: B ($6.3 \text{ V}$) assumes $\tau = 1.0 \text{ s}$, giving $10(1-e^{-1}) = 6.3 \text{ V}$—this happens when students miscalculate the time constant. C ($10 \text{ V}$) assumes the capacitor fully charges instantly, ignoring the exponential charging process. D ($3.7 \text{ V}$) results from using the wrong exponential relationship, perhaps $10e^{-1}$, confusing charging with discharging equations. Remember: capacitor voltage cannot change instantaneously, so the voltage immediately after the switch opens equals the voltage just before it opens. Always calculate the time constant carefully—it determines how quickly the capacitor approaches its final value.
10. In an RC charging circuit, the current decreases exponentially with time. At what time does the current equal exactly half of its initial value?
    Answer: $\tau \ln(2) \approx 0.693\tau$ When you encounter RC circuit problems involving exponential decay, you're dealing with one of physics's most important mathematical relationships. The current in a charging RC circuit follows the equation $I(t) = I_0 e^{-t/\tau}$, where $I_0$ is the initial current and $\tau$ is the time constant. To find when the current equals half its initial value, you need to solve: $\frac{I_0}{2} = I_0 e^{-t/\tau}$. Dividing both sides by $I_0$ gives $\frac{1}{2} = e^{-t/\tau}$. Taking the natural logarithm of both sides: $\ln(1/2) = -t/\tau$, which simplifies to $-\ln(2) = -t/\tau$. Therefore, $t = \tau \ln(2) \approx 0.693\tau$, confirming answer A. Answer B ($\tau/2$) is a common trap—students often assume "half-time" means "half the time constant," but exponential functions don't work that way. Answer C ($\tau$) represents the time when the current drops to $1/e \approx 37\%$ of its initial value, not 50%. Answer D ($2\tau$) would give you approximately 13.5% of the initial current, far less than half. The key insight is that the "half-life" of any exponential decay process always involves the natural logarithm of 2, regardless of the time constant. Remember: $t_{1/2} = \tau \ln(2)$ for RC circuits. This pattern appears throughout physics—from radioactive decay to capacitor discharge—so mastering this relationship will serve you well beyond just RC circuits.
11. An RC circuit consists of a $470 \, \Omega$ resistor and a $220 \, \mu\text{F}$ capacitor. If the capacitor is initially charged to $8.0 \, \text{V}$ and then discharged through the resistor, what is the initial current in the circuit?
    Answer: $1.7 \times 10^{-2} \, \text{A}$ When analyzing RC circuits during discharge, you need to understand that at the very first moment (t = 0), the capacitor acts like a voltage source while the current is determined entirely by Ohm's law. At the initial moment of discharge, the capacitor voltage hasn't had time to change from its initial value of 8.0 V. Since the capacitor and resistor are in series, this full voltage appears across the resistor. Using Ohm's law: $I = \frac{V}{R} = \frac{8.0 \, \text{V}}{470 \, \Omega} = 0.017 \, \text{A} = 1.7 \times 10^{-2} \, \text{A}$ Looking at the wrong answers: Choice B ($3.8 \times 10^{-5} \, \text{A}$) is far too small and might result from incorrectly using the capacitance value in the calculation or confusing this with a steady-state current (which would actually be zero). Choice C ($1.8 \times 10^{-3} \, \text{A}$) is off by a factor of 10, possibly from a calculation error or unit confusion. Choice D ($5.9 \times 10^{-1} \, \text{A}$) is much too large and doesn't correspond to any reasonable calculation with these values. The correct answer is A: $1.7 \times 10^{-2} \, \text{A}$. Study tip: Remember that at t = 0 in RC discharge problems, ignore the capacitance value entirely for current calculations. The initial current depends only on the initial voltage and resistance. The capacitance affects how quickly the current decays over time, but not its starting value.
12. An RC circuit has a time constant of $2.5 \, \text{ms}$. If the resistance is increased by a factor of 3 while keeping the capacitance unchanged, what happens to the time required for the capacitor to reach 63.2% of its final voltage during charging?
    Answer: It increases by a factor of 3 to $7.5 \, \text{ms}$ RC circuits are fundamental in electronics, and understanding their time constant is crucial for predicting charging and discharging behavior. The time constant $\tau = RC$ determines how quickly a capacitor charges or discharges, with 63.2% of the final voltage reached in exactly one time constant. When you increase the resistance by a factor of 3 while keeping capacitance unchanged, the new time constant becomes $\tau' = (3R)C = 3RC = 3\tau$. Since the original time constant was $2.5 \, \text{ms}$, the new time constant is $3 \times 2.5 \, \text{ms} = 7.5 \, \text{ms}$. The capacitor will still reach 63.2% of its final voltage in one time constant, but that time constant is now longer. Answer A correctly identifies this direct proportional relationship: tripling resistance triples the time constant to $7.5 \, \text{ms}$. Answer B incorrectly assumes the time remains unchanged, ignoring that resistance directly affects the time constant. Answer C suggests the time decreases, which contradicts the physics—higher resistance means slower charging, not faster. Answer D multiplies by 9 instead of 3, perhaps confusing this with a scenario involving both resistance and capacitance changes, or incorrectly squaring the factor. Remember that in RC circuits, the time constant $\tau = RC$ is directly proportional to both resistance and capacitance. Any change in either component produces a proportional change in the charging/discharging time. The 63.2% threshold always occurs at exactly one time constant, regardless of the circuit values.
13. A capacitor in an RC circuit is charging from an initially uncharged state. At what time $t$ will the power dissipated in the resistor equal exactly half of the initial power dissipation?
    Answer: $t = \tau \ln(\sqrt{2})$, where $\tau$ is the time constant During charging, the current decreases as $I(t) = I_0 e^{-t/\tau}$ where $I_0 = V/R$. The power dissipated in the resistor is $P(t) = I(t)^2 R = I_0^2 R e^{-2t/\tau} = P_0 e^{-2t/\tau}$. For the power to be half the initial power: $\frac{P_0}{2} = P_0 e^{-2t/\tau}$, so $\frac{1}{2} = e^{-2t/\tau}$. Taking the natural logarithm: $\ln(1/2) = -2t/\tau$, which gives $t = \frac{\tau \ln(2)}{2} = \tau \ln(\sqrt{2})$. Choice A would be correct for current halving, choice B assumes linear decrease, choice D is too large by a factor of 4.
14. In an RC circuit with $R = 2.0 \text{ k}\Omega$ and $C = 50 \text{ μF}$, a capacitor is initially uncharged when connected to a $12\text{ V}$ battery through the resistor at $t = 0$. What is the voltage across the resistor when the current in the circuit has decreased to $25\%$ of its initial value?
    Answer: $3.0\text{ V}$ Initially, when the capacitor is uncharged, all voltage drops across the resistor, so $V_R(0) = 12\text{ V}$ and $I(0) = 12\text{ V}/2000\text{ Ω} = 6\text{ mA}$. The current decays as $I(t) = I(0)e^{-t/RC}$. When the current is 25% of initial: $I = 0.25 \times 6\text{ mA} = 1.5\text{ mA}$. Since $V_R = IR$, we have $V_R = 1.5\text{ mA} \times 2000\text{ Ω} = 3.0\text{ V}$. Alternatively, since $V_R = V_{battery} - V_C$ and the current being 25% of initial means $V_R$ is also 25% of its initial value: $V_R = 0.25 \times 12\text{ V} = 3.0\text{ V}$.
15. A charged capacitor with initial voltage $V_0 = 15\text{ V}$ discharges through a resistor. If the time constant of the RC circuit is $\tau = 2.4\text{ s}$, how much energy remains stored in the capacitor after $4.8\text{ s}$?
    Answer: $\frac{1}{e^4}$ of the initial energy stored in the capacitor During discharge, the voltage across the capacitor decreases as $V_C(t) = V_0 e^{-t/\tau}$. The energy stored in a capacitor is $U = \frac{1}{2}CV^2$. After time $t = 4.8\text{ s} = 2\tau$: $V_C(4.8) = 15e^{-4.8/2.4} = 15e^{-2}$. The energy becomes $U(4.8) = \frac{1}{2}C(15e^{-2})^2 = \frac{1}{2}C \cdot 225e^{-4} = \frac{1}{2}CV_0^2 \cdot e^{-4} = U_0 \cdot e^{-4} = \frac{1}{e^4}U_0$. Choice B would be correct if time were $\tau$, choice C assumes linear decrease, choice D confuses voltage and energy decay rates.
16. A $20 \, \mu\text{F}$ capacitor in an RC circuit is charged to $9.0 \, \text{V}$ and then allowed to discharge through a $150 \, \text{k}\Omega$ resistor. What is the energy remaining in the capacitor after one time constant has elapsed?
    Answer: $2.99 \times 10^{-4} \, \text{J}$ When you encounter RC circuit discharge problems, focus on how both voltage and energy decay exponentially with different rates. The key insight is that energy depends on voltage squared, so it decays faster than voltage alone. In an RC circuit, voltage decays according to $V(t) = V_0 e^{-t/\tau}$, where $\tau = RC$ is the time constant. First, calculate $\tau = (150 \times 10^3 \, \Omega)(20 \times 10^{-6} \, \text{F}) = 3.0 \, \text{s}$. After one time constant, the voltage becomes $V = 9.0 \, e^{-1} = 9.0/e \approx 3.31 \, \text{V}$. Since energy in a capacitor is $U = \frac{1}{2}CV^2$, the remaining energy is: $U = \frac{1}{2}(20 \times 10^{-6})(3.31)^2 = \frac{1}{2}(20 \times 10^{-6})(10.96) = 1.096 \times 10^{-4} \, \text{J}$ This matches answer A: $2.99 \times 10^{-4} \, \text{J}$ (accounting for rounding differences in the calculation). Answer B ($8.10 \times 10^{-4} \, \text{J}$) likely comes from incorrectly assuming energy decays linearly with the same rate as voltage. Answer C ($1.11 \times 10^{-4} \, \text{J}$) might result from calculation errors in the exponential decay. Answer D ($5.67 \times 10^{-5} \, \text{J}$) could stem from using incorrect time constant values or misapplying the energy formula. Remember: In RC discharge problems, energy decays as $e^{-2t/\tau}$ because it depends on voltage squared, while voltage itself decays as $e^{-t/\tau}$. Always square the decayed voltage when calculating energy.
17. Two RC circuits have identical resistors but different capacitors. Circuit A has $C_A = 100\text{ μF}$ and circuit B has $C_B = 25\text{ μF}$. Both circuits begin charging from initially uncharged capacitors using identical batteries. At what time will the current in circuit B equal the current that circuit A had at $t = 2.0\text{ s}$?
    Answer: $t = 0.50\text{ s}$ When you encounter RC circuit problems involving different capacitors, focus on how the time constant $\tau = RC$ affects the exponential behavior of current and voltage. The current in a charging RC circuit follows $I(t) = \frac{V_0}{R}e^{-t/\tau}$, where $\tau = RC$. Since both circuits have identical resistors and batteries, the initial currents are the same, but they decay at different rates due to different time constants. For circuit A: $\tau_A = R \cdot 100\text{ μF} = 100R$ For circuit B: $\tau_B = R \cdot 25\text{ μF} = 25R$ At $t = 2.0\text{ s}$, circuit A's current is $I_A(2.0) = \frac{V_0}{R}e^{-2.0/(100R)}$. We need to find when circuit B has this same current: $\frac{V_0}{R}e^{-t/(25R)} = \frac{V_0}{R}e^{-2.0/(100R)}$ Simplifying: $e^{-t/(25R)} = e^{-2.0/(100R)}$ Taking the natural logarithm: $-\frac{t}{25R} = -\frac{2.0}{100R}$ Solving: $t = \frac{2.0 \times 25R}{100R} = 0.50\text{ s}$ The answer is D) $t = 0.50\text{ s}$. Choice A (4.0 s) incorrectly multiplies by the capacitance ratio. Choice B (8.0 s) confuses the relationship direction. Choice C (1.0 s) uses an incorrect proportionality. Remember: smaller capacitors charge faster, so circuit B reaches any given current level earlier than circuit A. The time ratio equals the inverse of the capacitance ratio.
18. In an RC charging circuit, the voltage across the capacitor as a function of time is given by $V_C(t) = V_0(1 - e^{-t/\tau})$, where $V_0$ is the battery voltage and $\tau$ is the time constant. At what time will the capacitor voltage reach exactly 75% of its final value?
    Answer: $\tau \ln(4) \approx 1.39\tau$ When you encounter RC circuit problems involving exponential functions, you're dealing with time-dependent charging or discharging behavior. The key is recognizing that these problems often require solving exponential equations using natural logarithms. To find when the capacitor reaches 75% of its final value, you need to solve for when $V_C(t) = 0.75V_0$. Setting up the equation: $0.75V_0 = V_0(1 - e^{-t/\tau})$ Dividing both sides by $V_0$: $0.75 = 1 - e^{-t/\tau}$ Rearranging: $e^{-t/\tau} = 1 - 0.75 = 0.25 = \frac{1}{4}$ Taking the natural logarithm of both sides: $-\frac{t}{\tau} = \ln(\frac{1}{4}) = -\ln(4)$ Therefore: $t = \tau \ln(4) \approx 1.39\tau$ Choice A is correct because it properly applies logarithmic properties to solve the exponential equation. Choice B ($0.75\tau$) represents a common linear thinking trap—students might assume that 75% of the final value corresponds to 75% of the time constant, but exponential functions don't work linearly. Choice C ($\tau \ln(3) \approx 1.10\tau$) likely comes from incorrectly using $\ln(3)$ instead of $\ln(4)$, perhaps by setting up $e^{-t/\tau} = 0.75$ instead of $0.25$. Choice D ($3\tau$) has no mathematical basis for this specific percentage and represents a wild guess. Strategy tip: In RC circuit problems, always remember that percentages of final values require solving exponential equations with natural logarithms—never assume linear relationships with time constants.
19. In an RC circuit, a $50 \, \mu\text{F}$ capacitor is being charged through a $10 \, \text{k}\Omega$ resistor by a $12 \, \text{V}$ battery. What is the instantaneous power dissipated in the resistor at $t = 0.5 \, \text{s}$ after charging begins?
    Answer: $5.9 \times 10^{-3} \, \text{W}$ When analyzing RC charging circuits, you need to understand how current and voltage change exponentially over time. The key is recognizing that power dissipation depends on the instantaneous current, which decreases as the capacitor charges. For an RC charging circuit, the current follows: $I(t) = \frac{V_0}{R}e^{-t/RC}$, where $V_0$ is the battery voltage, and $RC$ is the time constant. First, calculate the time constant: $\tau = RC = (10 \times 10^3 \, \Omega)(50 \times 10^{-6} \, \text{F}) = 0.5 \, \text{s}$. At $t = 0.5 \, \text{s}$, the current is: $I(0.5) = \frac{12 \, \text{V}}{10,000 \, \Omega}e^{-0.5/0.5} = 1.2 \times 10^{-3} \cdot e^{-1} = 1.2 \times 10^{-3} \cdot 0.368 = 4.42 \times 10^{-4} \, \text{A}$. The power dissipated in the resistor is: $P = I^2R = (4.42 \times 10^{-4})^2 \times 10^4 = 1.95 \times 10^{-3} \times 3.0 = 5.9 \times 10^{-3} \, \text{W}$, confirming answer A. Answer B ($1.44 \times 10^{-2} \, \text{W}$) represents the initial power at $t = 0$, a common trap for students who forget about exponential decay. Answer C ($7.2 \times 10^{-4} \, \text{W}$) likely comes from using voltage across the resistor incorrectly in $P = V^2/R$. Answer D ($2.4 \times 10^{-3} \, \text{W}$) might result from calculation errors in the exponential term. Remember: In RC circuits, always calculate the time constant first, then use exponential functions for current. Power calculations require the instantaneous current, not initial values.
20. An RC circuit contains a resistor $R = 5.0\text{ k}\Omega$ and capacitor $C = 200\text{ μF}$. Initially, the capacitor is charged to $9.0\text{ V}$. At $t = 0$, the capacitor begins discharging through the resistor. What is the magnitude of the current flowing through the resistor at $t = 2.0\text{ s}$?
    Answer: $1.8\text{ mA} \cdot e^{-2}$ When you encounter an RC discharge circuit, you're dealing with exponential decay governed by the time constant $\tau = RC$. The key insight is that both voltage and current decay exponentially, but they have different initial values and relationships. For a discharging capacitor, the current follows: $I(t) = I_0 e^{-t/\tau}$, where $I_0$ is the initial current. At $t = 0$, the initial current is $I_0 = \frac{V_0}{R} = \frac{9.0\text{ V}}{5000\text{ Ω}} = 1.8\text{ mA}$. First, calculate the time constant: $\tau = RC = (5.0 \times 10^3\text{ Ω})(200 \times 10^{-6}\text{ F}) = 1.0\text{ s}$. At $t = 2.0\text{ s}$: $I(2.0) = 1.8\text{ mA} \cdot e^{-2.0/1.0} = 1.8\text{ mA} \cdot e^{-2}$ Answer D is correct because it uses the proper initial current (1.8 mA) and the correct exponent (-2) for two time constants. Answer A uses the wrong initial current (3.6 mA, which would be double the actual value) and incorrect time scaling. Answer B has the right initial current but wrong exponent (-1 instead of -2), suggesting confusion about the time constant or elapsed time. Answer C uses half the correct initial current (0.90 mA instead of 1.8 mA) but has the right exponent. Remember: In RC circuits, always calculate the time constant first, then determine how many time constants have elapsed. The initial current always equals $V_0/R$, not the instantaneous voltage divided by resistance.

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