RC Circuits

RC Circuits — Part 1: RC Charging (Differential Equation)

An RC circuit consists of a resistor $R$ and a capacitor $C$ connected to a voltage source. When the switch closes, the capacitor charges through the resistor.

Setting up the differential equation

By Kirchhoff's voltage law around the loop:

$\mathcal{E} - IR - \frac{Q}{C} = 0$

Since $I = dQ/dt$:

$\mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0$

$\boxed{R\frac{dQ}{dt} + \frac{Q}{C} = \mathcal{E}}$

This is a first-order linear ODE — one of the most important differential equations in physics.

Solving the differential equation

Rearrange: $\dfrac{dQ}{dt} = \dfrac{\mathcal{E} - Q/C}{R} = \dfrac{C\mathcal{E} - Q}{RC}$

Separate variables:

$\frac{dQ}{C\mathcal{E} - Q} = \frac{dt}{RC}$

Integrate both sides (with $Q(0) = 0$):

$-\ln(C\mathcal{E} - Q)\Big|_0^Q = \frac{t}{RC}$

$-\ln\frac{C\mathcal{E} - Q}{C\mathcal{E}} = \frac{t}{RC}$

$C\mathcal{E} - Q = C\mathcal{E}\,e^{-t/(RC)}$

$\boxed{Q(t) = C\mathcal{E}\left(1 - e^{-t/(RC)}\right) = Q_{\max}\left(1 - e^{-t/\tau}\right)}$

where $\tau = RC$ is the time constant and $Q_{\max} = C\mathcal{E}$.

Current during charging

$I(t) = \frac{dQ}{dt} = \frac{\mathcal{E}}{R}e^{-t/(RC)} = I_0\,e^{-t/\tau}$

The current starts at $I_0 = \mathcal{E}/R$ and decays exponentially to zero.

Voltage across the capacitor

$V_C(t) = \frac{Q(t)}{C} = \mathcal{E}\left(1 - e^{-t/\tau}\right)$

Voltage across the resistor

$V_R(t) = IR = \mathcal{E}\,e^{-t/\tau}$

Check: $V_R + V_C = \mathcal{E}\,e^{-t/\tau} + \mathcal{E}(1 - e^{-t/\tau}) = \mathcal{E}$ ✓ (KVL satisfied at all times).

Key values

Time	$Q/Q_{\max}$	$V_C/\mathcal{E}$	$I/I_0$
$t = 0$	$0$	$0$	$1$
$t = \tau$	$1 - e^{-1} \approx 0.632$	$0.632$	$0.368$
$t = 2\tau$	$0.865$	$0.865$	$0.135$
$t = 5\tau$	$0.993$	$0.993$	$0.007$

Part 1 Summary

Quantity	Charging formula
Charge	$Q(t) = C\mathcal{E}(1 - e^{-t/\tau})$
Current	$I(t) = (\mathcal{E}/R)e^{-t/\tau}$
Capacitor voltage	$V_C(t) = \mathcal{E}(1 - e^{-t/\tau})$
Resistor voltage	$V_R(t) = \mathcal{E}\,e^{-t/\tau}$
Time constant	$\tau = RC$
ODE	$R\,dQ/dt + Q/C = \mathcal{E}$

Key insight: The RC charging equation is solved by separation of variables. The capacitor asymptotically approaches $V_C = \mathcal{E}$, while the current decays exponentially from $\mathcal{E}/R$ to zero.

RC Circuits — Part 2: RC Discharging

When a charged capacitor (initial charge $Q_0$) discharges through a resistor with no external EMF:

Kirchhoff's voltage law

$-IR - \frac{Q}{C} = 0 \implies R\frac{dQ}{dt} + \frac{Q}{C} = 0$

This is the homogeneous version of the charging ODE (no driving term $\mathcal{E}$).

Solution

$\frac{dQ}{Q} = -\frac{dt}{RC}$

$\ln Q - \ln Q_0 = -\frac{t}{RC}$

$\boxed{Q(t) = Q_0\,e^{-t/\tau}}$

where $\tau = RC$.

Current and voltage during discharge

Current

$I(t) = -\frac{dQ}{dt} = \frac{Q_0}{RC}e^{-t/\tau} = I_0\,e^{-t/\tau}$

where $I_0 = Q_0/(RC) = V_0/R$ (the minus sign is absorbed by choosing $I$ as the magnitude of current flow).

Capacitor voltage

$V_C(t) = \frac{Q(t)}{C} = \frac{Q_0}{C}e^{-t/\tau} = V_0\,e^{-t/\tau}$

Resistor voltage

$V_R(t) = IR = V_0\,e^{-t/\tau}$

During discharge, $V_R = V_C$ at all times (they are in a simple loop).

Comparison: Charging vs. Discharging

Quantity	Charging	Discharging
$Q(t)$	$Q_{\max}(1 - e^{-t/\tau})$	$Q_0\,e^{-t/\tau}$
$I(t)$	$I_0\,e^{-t/\tau}$	$I_0\,e^{-t/\tau}$
$V_C(t)$	$\mathcal{E}(1 - e^{-t/\tau})$	$V_0\,e^{-t/\tau}$

Notice: current always decays exponentially in both cases!

Half-life of an RC circuit

The half-life $t_{1/2}$ is the time for $Q$ (or $V_C$) to drop to half its initial value:

$Q_0/2 = Q_0\,e^{-t_{1/2}/\tau}$

$e^{-t_{1/2}/\tau} = 1/2$

$t_{1/2} = \tau \ln 2 \approx 0.693\tau$

Useful relationships

Elapsed time	Fraction remaining
$t_{1/2} \approx 0.693\tau$	$50\%$
$\tau$	$36.8\%$
$2\tau$	$13.5\%$
$3\tau$	$5.0\%$
$5\tau$	$0.67\%$

After $5\tau$, less than $1\%$ remains — the discharge is essentially complete.

Part 2 Summary

Quantity	Discharging formula
Charge	$Q(t) = Q_0\,e^{-t/\tau}$
Current	$I(t) = (V_0/R)\,e^{-t/\tau}$
Capacitor voltage	$V_C(t) = V_0\,e^{-t/\tau}$
ODE	$R\,dQ/dt + Q/C = 0$
Half-life	$t_{1/2} = \tau\ln 2 \approx 0.693\tau$

Key insight: Discharging is a simpler equation (homogeneous ODE). The solution is pure exponential decay — every quantity ($Q$, $I$, $V$) follows $e^{-t/\tau}$.

RC Circuits — Part 3: Time Constant τ = RC

The time constant $\tau = RC$ is the single most important parameter of an RC circuit. It sets the timescale for all exponential behavior.

Physical meaning

$\tau = RC$

$R$ large	Charges flow slowly → slow charging/discharging
$C$ large	More charge to move → slow charging/discharging
Both large	Very slow response
Both small	Very fast response

Units verification

$[\tau] = [R][C] = \Omega \cdot \text{F} = \frac{\text{V}}{\text{A}} \cdot \frac{\text{C}}{\text{V}} = \frac{\text{C}}{\text{A}} = \frac{\text{C}}{\text{C/s}} = \text{s} \checkmark$

What happens at $t = \tau$?

During charging ($\mathcal{E}$ applied):

• $Q(\tau) = Q_{\max}(1 - e^{-1}) \approx 0.632\,Q_{\max}$
• Capacitor is $63.2\%$ charged
• Current has dropped to $36.8\%$ of $I_0$

During discharging:

• $Q(\tau) = Q_0\,e^{-1} \approx 0.368\,Q_0$
• Capacitor retains $36.8\%$ of its charge
• $63.2\%$ of the charge has flowed out

Slope interpretation

At $t = 0$ during discharge, the tangent line to $Q(t)$ has slope:

$\left.\frac{dQ}{dt}\right|_{t=0} = -\frac{Q_0}{\tau}$

This tangent line would reach $Q = 0$ at $t = \tau$. In other words, if the initial rate of discharge continued unchanged, the capacitor would fully discharge in time $\tau$.

Designing with time constants

Example: Delay circuit

Want a $2$ s delay before a voltage reaches $70\%$ of $\mathcal{E}$:

$0.70 = 1 - e^{-2/\tau} \implies e^{-2/\tau} = 0.30 \implies \tau = \frac{2}{\ln(1/0.30)} = \frac{2}{1.204} \approx 1.66 \text{ s}$

Choose $R$ and $C$ such that $RC = 1.66$ s. For example: $R = 166$ kΩ, $C = 10$ μF.

Multiple RC stages

For a circuit with several resistors and one capacitor, use the equivalent resistance seen by the capacitor:

$\tau = R_{\text{eq}} \cdot C$

where $R_{\text{eq}}$ is the resistance between the capacitor's terminals with the ideal battery replaced by a wire and $C$ removed.

Part 3 Summary

Property	Value
Time constant	$\tau = RC$
Units	seconds
At $t = \tau$ (charging)	$63.2\%$ charged
At $t = \tau$ (discharging)	$36.8\%$ remaining
Time to reach fraction $f$	$t = -\tau\ln(1-f)$ (charging)
Half-life	$t_{1/2} = \tau\ln 2 \approx 0.693\tau$
Several resistors	$\tau = R_{\text{eq}} C$ (resistance seen by $C$)

Key insight: $\tau = RC$ controls everything. On the AP exam, if they give you $R$ and $C$, compute $\tau$ immediately — it's almost certainly needed.

RC Circuits — Part 4: Current and Voltage Graphs

Understanding the shapes of RC circuit graphs is essential for the AP exam. Every graph is built from two patterns: exponential growth ($1 - e^{-t/\tau}$) and exponential decay ($e^{-t/\tau}$).

The two fundamental shapes

Exponential decay $f(t) = A\,e^{-t/\tau}$:

• Starts at $A$, decreases toward $0$
• Steepest at $t = 0$
• Concave up (curves upward)

Exponential growth $f(t) = A(1 - e^{-t/\tau})$:

• Starts at $0$, increases toward $A$
• Steepest at $t = 0$
• Concave down (curves downward)

Charging graphs

Capacitor voltage $V_C$

$V_C(t) = \mathcal{E}(1 - e^{-t/\tau})$

• Shape: exponential growth (concave down)
• $V_C(0) = 0$, $V_C(\infty) = \mathcal{E}$
• Slope at $t = 0$: $dV_C/dt|_0 = \mathcal{E}/\tau$

Current $I$

$I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}$

• Shape: exponential decay (concave up)
• $I(0) = \mathcal{E}/R$, $I(\infty) = 0$

Resistor voltage $V_R$

$V_R(t) = \mathcal{E}\,e^{-t/\tau}$

• Same shape as current (exponential decay)
• $V_R = IR$, so $V_R$ is proportional to $I$

Check: $V_R + V_C = \mathcal{E}$ at all times

At any instant, $V_R$ and $V_C$ are complementary: they add up to $\mathcal{E}$. If you flip the $V_C$ graph upside down and shift it, you get $V_R$.

Discharging graphs

All quantities decay exponentially:

$Q(t) = Q_0\,e^{-t/\tau}, \quad V_C(t) = V_0\,e^{-t/\tau}, \quad I(t) = I_0\,e^{-t/\tau}$

All three have the same shape: exponential decay (concave up), starting at their maximum values and approaching zero.

Time to reach a specific value

To find when $V_C$ reaches a target $V_{\text{target}}$:

$V_{\text{target}} = V_0\,e^{-t/\tau}$

$t = -\tau\ln\left(\frac{V_{\text{target}}}{V_0}\right) = \tau\ln\left(\frac{V_0}{V_{\text{target}}}\right)$

Slope at $t = 0$ (discharge)

$\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{\tau}$

The initial slope is steepest and equals $-V_0/\tau$. A tangent line at $t = 0$ crosses zero at $t = \tau$.

Switched RC circuits

A common AP problem: a capacitor charges for a while, then the switch redirects to discharge through a different resistor.

Example

Charge through $R_1 = 10$ kΩ with $\mathcal{E} = 20$ V, $C = 50$ μF for $t_1 = 1$ s.

$\tau_1 = R_1 C = 0.5$ s. After $t_1/\tau_1 = 2$ time constants:

$V_C = 20(1 - e^{-2}) \approx 20(0.865) = 17.3 \text{ V}$

Then discharge through $R_2 = 20$ kΩ. New time constant: $\tau_2 = R_2 C = 1.0$ s.

$V_C(t) = 17.3\,e^{-t/1.0}$

The graph shows a kink at the switching time — the voltage is continuous but the slope changes because $\tau$ changes.

Part 4 Summary

Graph	Charging	Discharging
$V_C$	Growth: $\mathcal{E}(1-e^{-t/\tau})$ ↗	Decay: $V_0 e^{-t/\tau}$ ↘
$I$	Decay: $(\mathcal{E}/R)e^{-t/\tau}$ ↘	Decay: $(V_0/R)e^{-t/\tau}$ ↘
$V_R$	Decay: $\mathcal{E}\,e^{-t/\tau}$ ↘	Decay: $V_0 e^{-t/\tau}$ ↘
Shape (decay)	Concave up	Concave up
Shape (growth)	Concave down	—

Exam tip: You'll often be asked to sketch or identify these graphs. Remember: the current always decays exponentially in RC circuits. Only $V_C$ during charging shows exponential growth.

RC Circuits — Part 5: Power and Energy in RC Circuits

Energy conservation in RC circuits connects to calculus through integration of power over time.

Power dissipated in the resistor

$P_R(t) = I^2 R$

During charging

$P_R(t) = \frac{\mathcal{E}^2}{R}e^{-2t/\tau}$

Note the factor of $2$ in the exponent: since $P \propto I^2$, the power decays twice as fast as the current.

During discharging

$P_R(t) = \frac{V_0^2}{R}e^{-2t/\tau}$

Total energy dissipated during charging

$W_R = \int_0^\infty P_R\,dt = \int_0^\infty \frac{\mathcal{E}^2}{R}e^{-2t/\tau}\,dt$

$= \frac{\mathcal{E}^2}{R}\left[-\frac{\tau}{2}e^{-2t/\tau}\right]_0^\infty = \frac{\mathcal{E}^2}{R}\cdot\frac{\tau}{2} = \frac{\mathcal{E}^2 RC}{2R} = \frac{1}{2}C\mathcal{E}^2$

Energy budget during charging

Destination	Energy	Fraction
Stored in capacitor	$\frac{1}{2}C\mathcal{E}^2$	$50\%$
Dissipated in $R$	$\frac{1}{2}C\mathcal{E}^2$	$50\%$
Supplied by battery	$C\mathcal{E}^2$	$100\%$

Remarkable result: Exactly half the energy goes to the capacitor and half to the resistor — regardless of $R$! A larger $R$ means slower charging and lower current, but the same total heat.

Energy during discharging

During discharge, all the capacitor's energy is dissipated in $R$:

$W_R = \int_0^\infty \frac{V_0^2}{R}e^{-2t/\tau}\,dt = \frac{V_0^2}{R}\cdot\frac{\tau}{2} = \frac{V_0^2 C}{2} = \frac{1}{2}CV_0^2 = U_0$

This confirms energy conservation: the initial stored energy $U_0 = \frac{1}{2}CV_0^2$ is entirely converted to heat.

Power delivered by the battery (charging)

$P_{\text{batt}} = \mathcal{E} I = \frac{\mathcal{E}^2}{R}e^{-t/\tau}$

Total energy delivered:

$W_{\text{batt}} = \int_0^\infty \frac{\mathcal{E}^2}{R}e^{-t/\tau}\,dt = \frac{\mathcal{E}^2}{R}\cdot\tau = \frac{\mathcal{E}^2 C}{1} \cdot 1 = C\mathcal{E}^2$

Note: $P_{\text{batt}}$ decays as $e^{-t/\tau}$ while $P_R$ decays as $e^{-2t/\tau}$—the battery power decays half as fast on a log scale.

Instantaneous energy stored in the capacitor

$U_C(t) = \frac{Q(t)^2}{2C}$

During charging:

$U_C(t) = \frac{C\mathcal{E}^2}{2}(1 - e^{-t/\tau})^2$

Rate of energy storage:

$\frac{dU_C}{dt} = \frac{Q}{C}\cdot\frac{dQ}{dt} = V_C \cdot I = \mathcal{E}(1-e^{-t/\tau})\cdot\frac{\mathcal{E}}{R}e^{-t/\tau}$

$= \frac{\mathcal{E}^2}{R}e^{-t/\tau}(1 - e^{-t/\tau})$

This rate is zero at $t = 0$ (no charge stored yet) and at $t = \infty$ (current is zero). It peaks at $t = \tau\ln 2$, when $V_C = V_R = \mathcal{E}/2$.

Part 5 Summary

Quantity	Expression
Power in $R$	$P_R = I^2 R = (V_0^2/R)e^{-2t/\tau}$
Energy dissipated (charge)	$W_R = \frac{1}{2}C\mathcal{E}^2$
Energy dissipated (discharge)	$W_R = \frac{1}{2}CV_0^2 = U_0$
Battery energy (charging)	$W_{\text{batt}} = C\mathcal{E}^2$
Efficiency	$50\%$ (always, regardless of $R$!)
Max $dU_C/dt$	at $t = \tau\ln 2$

Key insight: During charging, exactly half the battery's energy is dissipated as heat, regardless of resistance. This "50% rule" is a consequence of charging a capacitor ($V = Q/C$) from a constant-emf battery.

RC Circuits — Part 6: Problem-Solving Workshop

These problems cover the full range of RC circuit analysis expected on the AP Physics C: E&M exam, including differential equations, energy, and multi-component circuits.

Problem-solving checklist

1. Identify charging or discharging
2. Compute $\tau = RC$ (or $R_{\text{eq}}C$, using the resistance seen by the capacitor)
3. Write the appropriate equation ($Q$, $I$, or $V$)
4. Apply initial and boundary conditions
5. Integrate if energy/charge questions arise

Problem 2: Finding $R$ from a discharge curve

A capacitor $C = 10$ μF starts at $V_0 = 50$ V. After $3$ ms, $V = 6.77$ V. Find $R$.

Solution

$V = V_0 e^{-t/\tau}$

$6.77 = 50\,e^{-0.003/\tau}$

$e^{-0.003/\tau} = 0.1354$

$-0.003/\tau = \ln(0.1354) = -2.000$

$\tau = 0.003/2 = 0.0015 \text{ s} = 1.5 \text{ ms}$

$R = \tau/C = 0.0015/(10 \times 10^{-6}) = 150\;\Omega$

Problem 3: Two-resistor RC circuit

A circuit has $\mathcal{E} = 20$ V, $R_1 = 4$ kΩ (in series), and $R_2 = 6$ kΩ (in parallel with $C = 50$ μF).

(a) Long-time behavior ($t \to \infty$)

When fully charged, $I_C = 0$. All current flows through $R_1$ and $R_2$:

$I_{\infty} = \frac{\mathcal{E}}{R_1 + R_2} = \frac{20}{10\,000} = 2 \text{ mA}$

$V_C = I_{\infty} R_2 = 2 \times 6 = 12 \text{ V}$

(b) Time constant

Zero the source, remove $C$: from $C$'s terminals, $R_1$ and $R_2$ are in parallel:

$R_{\text{eq}} = R_1 \| R_2 = \frac{4 \times 6}{10} = 2.4 \text{ kΩ}$

$\tau = R_{\text{eq}} C = 2400 \times 50 \times 10^{-6} = 0.12 \text{ s}$

(c) Capacitor voltage

$V_C(t) = 12(1 - e^{-t/0.12})$

Problem 4: Charge transfer between capacitors

$C_1 = 10$ μF is charged to $V_1 = 100$ V. It is then connected to uncharged $C_2 = 40$ μF through $R = 1$ kΩ.

Differential equation

$\frac{Q_1}{C_1} = \frac{Q_2}{C_2} + I R$

With charge conservation: $Q_1 + Q_2 = Q_{1,0} = C_1 V_1 = 1$ mC.

Let $Q_2 = q$ (charge transferred). Then $Q_1 = Q_{1,0} - q$:

$\frac{Q_{1,0} - q}{C_1} - \frac{q}{C_2} = R\frac{dq}{dt}$

The effective time constant:

$\tau = R\cdot\frac{C_1 C_2}{C_1 + C_2} = R C_{\text{series}} = 1000 \times \frac{10 \times 40}{50} \times 10^{-6} = 8 \text{ ms}$

Final voltages: $V_f = Q_{1,0}/(C_1 + C_2) = 10^{-3}/(50 \times 10^{-6}) = 20$ V on both.

Part 6 Summary

Key problem types

Type	Approach
Basic charging/discharging	$Q(t)$, $I(t)$, $V(t)$ formulas
Find $R$ or $C$ from data	Solve $V = V_0 e^{-t/\tau}$ for $\tau$, then $R = \tau/C$
Multi-resistor circuits	Find $R_{\text{eq}}$ seen by $C$, then $\tau = R_{\text{eq}}C$
Capacitor-to-capacitor	Use charge conservation + series $C$ time constant
Energy problems	Integrate $P_R = I^2R$ or use energy conservation

Exam strategy: For free-response, always (1) write the differential equation, (2) state the solution, (3) verify initial/final conditions. Partial credit is awarded for each step.

RC Circuits — Part 7: Review & Applications

Complete formula reference

Quantity	Charging	Discharging
$Q(t)$	$C\mathcal{E}(1-e^{-t/\tau})$	$Q_0 e^{-t/\tau}$
$I(t)$	$(\mathcal{E}/R)e^{-t/\tau}$	$(V_0/R)e^{-t/\tau}$
$V_C(t)$	$\mathcal{E}(1-e^{-t/\tau})$	$V_0 e^{-t/\tau}$
$V_R(t)$	$\mathcal{E}\,e^{-t/\tau}$	$V_0 e^{-t/\tau}$

Differential equations

Scenario	ODE
Charging	$RC\,dQ/dt + Q = C\mathcal{E}$
Discharging	$RC\,dQ/dt + Q = 0$

Energy

Quantity	Expression
Battery delivers (charging)	$C\mathcal{E}^2$
Stored in $C$ (charging)	$\frac{1}{2}C\mathcal{E}^2$
Dissipated in $R$ (charging)	$\frac{1}{2}C\mathcal{E}^2$
Dissipated in $R$ (discharging)	$\frac{1}{2}CV_0^2$

Real-World Applications

Camera flash

A camera flash uses an RC charging circuit. The capacitor charges slowly from a small battery (large $\tau$), then discharges rapidly through the flash tube (small $R_{\text{flash}}$, small $\tau$).

Heart defibrillator

A defibrillator charges a large capacitor ($C \sim 30{-}70$ μF) to high voltage ($V \sim 1000{-}5000$ V), then discharges through the patient's chest ($R \sim 50$ Ω). Time constant $\sim 2{-}4$ ms matches the needed pulse duration.

Reading a discharge from data

If a capacitor's voltage falls from $V_0$ to $V$ in time $t$, then $t/\tau = \ln(V_0/V)$, so $R = t/\big(C\ln(V_0/V)\big)$. Plotting $\ln V$ against $t$ gives a straight line of slope $-1/\tau$ — a standard way to find $RC$ from lab data.

Common AP Exam Mistakes

Mistake	Correction
Using $V_C = \mathcal{E}e^{-t/\tau}$ for charging	That's discharging! Charging: $V_C = \mathcal{E}(1 - e^{-t/\tau})$
Forgetting $V_C$ can't jump	Capacitor voltage is always continuous
Using wrong $R$ for $\tau$	In complex circuits, use the equivalent resistance seen by $C$
Confusing $e^{-t/\tau}$ with $e^{-2t/\tau}$	Power goes as $I^2$, so $P \propto e^{-2t/\tau}$
Not setting up the ODE	On free response, always write KVL → ODE → solution
Wrong sign on $I = dQ/dt$	Define $I$ direction clearly; $I = +dQ/dt$ for charging

Free-response template

1. Draw the circuit and label $I$ direction
2. Write KVL: $\mathcal{E} - IR - Q/C = 0$
3. Substitute $I = dQ/dt$
4. Solve by separation of variables
5. Apply initial condition $Q(0) = 0$ or $Q(0) = Q_0$
6. Answer the specific question (find $t$, $V$, $I$, etc.)

Topic Complete: RC Circuits

You've mastered RC circuits for AP Physics C: E&M:

Part	Topic	Status
1	Charging differential equation	✅
2	Discharging	✅
3	Time constant $\tau = RC$	✅
4	Current & voltage graphs	✅
5	Power & energy	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Exam tip: RC circuits appear on nearly every AP Physics C: E&M exam. The differential equation, its solution, and the energy analysis are all fair game for free-response. Practice writing the full derivation from KVL → ODE → separation of variables → solution → initial conditions.