AP Physics C E&M- 11.8 Resistor-Capacitor (RC) Circuits- Study Notes- New Syllabus
AP Physics C E&M- 11.8 Resistor-Capacitor (RC) Circuits – Study Notes
AP Physics C E&M- 11.8 Resistor-Capacitor (RC) Circuits – Study Notes – per latest Syllabus.
Key Concepts:
- Equivalent Capacitance of Multiple Capacitors
- RC Circuits: Behavior of Resistor–Capacitor Combinations
- Transient and Steady-State Behavior of a Capacitor in a Circuit
Equivalent Capacitance of Multiple Capacitors
Equivalent capacitance is the single capacitance that can replace a combination of capacitors in a circuit without changing the overall charge–voltage relationship.
Capacitors in Parallel:
All capacitors are connected across the same potential difference \(V\).
Total charge stored is the sum of charges on each capacitor:
\( \mathrm{Q_{total} = Q_1 + Q_2 + Q_3 + \dots} \)
Since \( \mathrm{Q = C V} \), the equivalent capacitance is:
\( \mathrm{C_{eq} = C_1 + C_2 + C_3 + \dots} \)
Key Idea: Equivalent capacitance in parallel is larger than any individual capacitor.
Capacitors in Series:
Conservation of charge requires each capacitor in series to have the same charge magnitude.
Total potential difference is the sum of voltage drops across each capacitor:
\( \mathrm{V = V_1 + V_2 + V_3 + \dots} \)
Since \( \mathrm{V = \dfrac{Q}{C}} \), the reciprocal form of equivalent capacitance is:
\( \mathrm{\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \dots} \)
Key Idea: Equivalent capacitance in series is always smaller than the smallest individual capacitor.
Mixed Series–Parallel Combinations:
- For circuits with both series and parallel capacitors, simplify step by step.
- First reduce parallel groups, then reduce series (or vice versa) until a single equivalent capacitance remains.
Key Idea: Capacitors in parallel add directly, while capacitors in series combine inversely. Conservation of charge ensures equal charge on series capacitors, while conservation of potential difference ensures equal voltage across parallel capacitors.
Example
Two capacitors of \( 6.0 \, \mu F \) and \( 4.0 \, \mu F \) are connected in parallel across a \(12 \, V\) battery. Find the equivalent capacitance and the total charge stored.
▶️ Answer/Explanation
Step 1: Equivalent capacitance: \( \mathrm{C_{eq} = C_1 + C_2 = 6.0 + 4.0 = 10.0 \, \mu F} \).
Step 2: Total charge: \( \mathrm{Q = C_{eq} V = (10.0 \times 10^{-6})(12)} = 1.2 \times 10^{-4} \, C \).
Final Answer: \( \mathrm{C_{eq} = 10.0 \, \mu F} \), and total charge stored = \( \mathrm{1.2 \times 10^{-4} \, C} \).
Example
Two capacitors of \( 12 \, \mu F \) and \( 6.0 \, \mu F \) are connected in series across a \(9.0 \, V\) battery. Find the equivalent capacitance and the charge on each capacitor.
▶️ Answer/Explanation
Step 1: Equivalent capacitance: \( \mathrm{\dfrac{1}{C_{eq}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1}{12} + \dfrac{2}{12} = \dfrac{3}{12}} \).
\( \mathrm{C_{eq} = 4.0 \, \mu F} \).
Step 2: Total charge: \( \mathrm{Q = C_{eq} V = (4.0 \times 10^{-6})(9.0) = 3.6 \times 10^{-5} \, C} \).
Step 3: Charge on each capacitor is the same in series: \( Q = 3.6 \times 10^{-5} \, C \).
Step 4: Voltage across each capacitor: \( V_1 = Q/C_1 = (3.6 \times 10^{-5})/(12 \times 10^{-6}) = 3.0 \, V \). \( V_2 = Q/C_2 = (3.6 \times 10^{-5})/(6 \times 10^{-6}) = 6.0 \, V \).
Final Answer: \( \mathrm{C_{eq} = 4.0 \, \mu F} \), with capacitor voltages \( V_1 = 3.0 \, V \) and \( V_2 = 6.0 \, V \).
RC Circuits: Behavior of Resistor–Capacitor Combinations
An RC circuit is an electric circuit containing both resistors and capacitors. The charge on the capacitor and the current in the circuit change over time according to exponential functions derived from Kirchhoff’s loop rule.
Differential Equation (from Kirchhoff’s Loop Rule):
- For charging: \( \mathrm{\mathcal{E} – \dfrac{q}{C} – IR = 0} \), where \( I = \dfrac{dq}{dt} \). This leads to: \( \mathrm{R \dfrac{dq}{dt} + \dfrac{q}{C} = \mathcal{E}} \).
- For discharging: \( \mathrm{\dfrac{q}{C} + IR = 0} \). This leads to: \( \mathrm{R \dfrac{dq}{dt} + \dfrac{q}{C} = 0} \).
Time Constant:
\( \mathrm{\tau = RC} \)
- The time constant \(\tau\) sets the time scale for charging or discharging.
- After one time constant, a charging capacitor reaches about 63% of its final charge, and a discharging capacitor drops to about 37% of its initial charge.
Charging a Capacitor:
Charge as a function of time:
\( \mathrm{q(t) = C\mathcal{E}\left(1 – e^{-t/RC}\right)} \)
Current as a function of time:
\( \mathrm{I(t) = \dfrac{\mathcal{E}}{R} e^{-t/RC}} \)
As \( t \to \infty \), \( q \to C\mathcal{E} \) and \( I \to 0 \).
Discharging a Capacitor:
Charge as a function of time:
\( \mathrm{q(t) = q_0 e^{-t/RC}} \)
Current as a function of time:
\( \mathrm{I(t) = -\dfrac{q_0}{RC} e^{-t/RC}} \)
As \( t \to \infty \), both \( q \) and \( I \) approach 0.
Key Idea: The exponential charging and discharging behavior of an RC circuit is governed by the time constant \( \tau = RC \), which determines how quickly the circuit responds to changes.
Example
A \(10 \, \mu F\) capacitor is connected in series with a \(20 \, k\Omega\) resistor and a \(12 \, V\) battery. Find the time constant and the charge on the capacitor after \(1.0 \, s\).
▶️ Answer/Explanation
Step 1: Time constant: \( \mathrm{\tau = RC = (20 \times 10^3)(10 \times 10^{-6}) = 0.20 \, s} \).
Step 2: Final charge: \( \mathrm{q_{max} = C\mathcal{E} = (10 \times 10^{-6})(12) = 1.2 \times 10^{-4} \, C} \).
Step 3: Charge at \( t = 1.0 \, s \): \( \mathrm{q(t) = q_{max}(1 – e^{-t/\tau})} \).
\( \mathrm{q(1.0) = 1.2 \times 10^{-4}(1 – e^{-1.0/0.20})} \).
Step 4: \( e^{-5} \approx 0.0067 \). So \( \mathrm{q(1.0) \approx 1.2 \times 10^{-4}(0.9933) \approx 1.19 \times 10^{-4} \, C} \).
Final Answer: Time constant = \(0.20 \, s\). Charge after \(1.0 \, s\) = \(1.19 \times 10^{-4} \, C\) (nearly fully charged).
Example
A capacitor initially charged to \(6.0 \, \mu C\) is discharged through a \(3.0 \, k\Omega\) resistor. The capacitance is \(2.0 \, \mu F\). Find the time constant and the charge remaining after \(4.5 \, ms\).
▶️ Answer/Explanation
Step 1: Time constant: \( \mathrm{\tau = RC = (3000)(2.0 \times 10^{-6}) = 6.0 \times 10^{-3} \, s = 6.0 \, ms} \).
Step 2: Initial charge: \( \mathrm{q_0 = 6.0 \, \mu C} \).
Step 3: Charge after \(t = 4.5 \, ms\): \( \mathrm{q(t) = q_0 e^{-t/\tau} = 6.0 \times 10^{-6} e^{-4.5/6.0}} \).
Step 4: \( e^{-0.75} \approx 0.472 \). \( \mathrm{q(t) \approx 6.0 \times 10^{-6} (0.472) = 2.83 \times 10^{-6} \, C} \).
Final Answer: Time constant = \(6.0 \, ms\). Charge after \(4.5 \, ms\) = \(2.83 \, \mu C\).
Transient and Steady-State Behavior of a Capacitor in a Circuit
Initial Behavior of a Capacitor:
- Immediately after being connected to a circuit, an uncharged capacitor acts like a wire (short circuit).
- Charge flows easily to or from the plates of the capacitor, producing an initial current in the branch containing the capacitor.
Charging Process:
As the capacitor charges, the potential difference across it increases.
The charge on the plates grows according to:
\( \mathrm{q(t) = C \mathcal{E} \left(1 – e^{-t/RC}\right)} \)
The current in the branch decreases exponentially:
\( \mathrm{I(t) = \dfrac{\mathcal{E}}{R} e^{-t/RC}} \)
The energy stored in the capacitor increases with time:
\( \mathrm{U(t) = \tfrac{1}{2} C [V(t)]^2} \)
All of these quantities approach steady-state values asymptotically.
Steady State After Long Time (Charging):
- After many time constants (\(t \gg RC\)), the capacitor is fully charged.
- Potential difference across capacitor: \( V = \mathcal{E} \).
- Current in branch: \( I = 0 \) (open-circuit behavior).
- Stored energy: \( \mathrm{U = \tfrac{1}{2} C \mathcal{E}^2} \).
Discharging Process:
Immediately after discharging begins, the capacitor starts to release stored charge and energy.
The charge decreases exponentially:
\( \mathrm{q(t) = q_0 e^{-t/RC}} \)
The current decreases with time:
\( \mathrm{I(t) = -\dfrac{q_0}{RC} e^{-t/RC}} \)
The potential difference across the capacitor decreases with time:
\( \mathrm{V(t) = V_0 e^{-t/RC}} \)
The stored energy decreases as:
\( \mathrm{U(t) = \tfrac{1}{2} C [V_0 e^{-t/RC}]^2} \)
Steady State After Long Time (Discharging):
- After many time constants (\(t \gg RC\)), the capacitor becomes fully discharged.
- Charge: \( q = 0 \).
- Potential difference: \( V = 0 \).
- Current: \( I = 0 \).
- Stored energy: \( U = 0 \).
Key Idea: The potential difference, charge, current, and energy in an RC circuit all change exponentially with time. The time constant \( \tau = RC \) determines how quickly the capacitor transitions between its transient state and steady-state conditions.
Example
A \( 5.0 \, \mu F \) capacitor is connected in series with a \( 1.0 \, k\Omega \) resistor and a \( 12 \, V \) battery. Find the capacitor’s potential difference after \( 5.0 \, ms \).
▶️ Answer/Explanation
Step 1: Time constant: \( \mathrm{\tau = RC = (1000)(5.0 \times 10^{-6}) = 5.0 \, ms} \).
Step 2: Voltage across capacitor: \( \mathrm{V(t) = \mathcal{E} (1 – e^{-t/\tau})} \).
Step 3: At \(t = 5.0 \, ms\): \( \mathrm{V = 12(1 – e^{-1})} \).
\( e^{-1} \approx 0.368 \). \( \mathrm{V \approx 12(0.632) = 7.6 \, V} \).
Final Answer: After \( 5.0 \, ms \), the capacitor’s potential difference is approximately \( 7.6 \, V \).