AP Physics 2- 11.8 Resistor-Capacitor (RC) Circuits- Study Notes- New Syllabus
AP Physics 2- 11.8 Resistor-Capacitor (RC) Circuits – Study Notes
AP Physics 2- 11.8 Resistor-Capacitor (RC) Circuits – Study Notes – per latest Syllabus.
Key Concepts:
- Equivalent Capacitance of Multiple Capacitors
- RC Circuits (Resistor–Capacitor Circuits)
Equivalent Capacitance of Multiple Capacitors
The equivalent capacitance of a combination of capacitors is the single capacitance that can replace them in a circuit without changing the overall charge–voltage relationship.
Capacitors in Series:
The charge on each capacitor is the same: \( \rm {Q_1 = Q_2 = \dots = Q} \).
- The reciprocal of the total capacitance is the sum of reciprocals of individual capacitances:
\(\dfrac{1}{\text{C}_{\text{eq}}} =\rm{ \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots}\)
- The equivalent capacitance is always less than the smallest individual capacitance.
- Voltage across each capacitor adds up to the total applied voltage: \(\text{V}_{\text{total}} = V_1 + V_2 + \dots\).
- Charge–Voltage Rule: All capacitors have the same charge \(Q\), but individual voltages differ: \(V_i = \dfrac{\rm Q}{\rm {C_i}}\).
Capacitors in Parallel:
Voltage across each capacitor is the same: \( \rm {V_1 = V_2 = \dots = V} \).
- The total (equivalent) capacitance is the sum of all capacitances:
\(\text{C}_{\text{eq}} =\rm { C_1 + C_2 + \dots}\)
- The equivalent capacitance is always greater than the largest individual capacitance.
- Total charge is the sum of charges stored in each capacitor: \(\text{Q}_{\text{total}} =\rm { Q_1 + Q_2 + \dots}\).
- Charge–Voltage Rule: Each capacitor has the same voltage \(V\), but individual charges differ: \(\rm {Q_i = C_i V}\).
Combination Circuits:
- Circuits often contain both series and parallel parts.
- Apply the series formula first, then the parallel formula step by step until the circuit is simplified.
Example:
Two capacitors, \(C_1 = 6 \, \mu F\) and \(C_2 = 3 \, \mu F\), are connected in series to a \(12 \, V\) battery. Find the equivalent capacitance, the charge on each capacitor, and the voltage across each.
▶️ Answer/Explanation
Step 1: Series formula: \(\dfrac{1}{\text{C}_{\text{eq}}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{3}{6} \Rightarrow \text{C}_{\text{eq}} = 2 \, \mu F\).
Step 2: Total charge: \(Q = C_{\text{eq}} V = 2 \cdot 12 = 24 \, \mu C\).
Step 3: Voltage across each capacitor: \(V_1 = Q / C_1 = 24 / 6 = 4 \, V\), \(V_2 = Q / C_2 = 24 / 3 = 8 \, V\).
Answer: \(C_{\text{eq}} = 2 \, \mu F\), \(Q = 24 \, \mu C\), \(V_1 = 4 \, V\), \(V_2 = 8 \, V\).
Example :
Two capacitors, \(C_1 = 4 \, \mu F\) and \(C_2 = 5 \, \mu F\), are connected in parallel across a \(10 \, V\) battery. Find the equivalent capacitance, the charge on each capacitor, and the total charge.
▶️ Answer/Explanation
Step 1: Parallel formula: \(\text{C}_{\text{eq}} = C_1 + C_2 = 4 + 5 = 9 \, \mu F\).
Step 2: Total charge: \(Q_{\text{total}} = C_{\text{eq}} V = 9 \cdot 10 = 90 \, \mu C\).
Step 3: Charge on each capacitor: \(Q_1 = C_1 V = 4 \cdot 10 = 40 \, \mu C\), \(Q_2 = C_2 V = 5 \cdot 10 = 50 \, \mu C\).
Answer: \(C_{\text{eq}} = 9 \, \mu F\), \(Q_{\text{total}} = 90 \, \mu C\), \(Q_1 = 40 \, \mu C\), \(Q_2 = 50 \, \mu C\).
RC Circuits (Resistor–Capacitor Circuits)
An RC circuit is an electric circuit composed of a resistor and a capacitor connected in series or parallel. These circuits exhibit time-dependent behavior when a voltage is applied or removed due to the charging and discharging of the capacitor.
Time Constant (\( \tau \)):
- The time constant is a significant feature of an RC circuit that measures how quickly a capacitor will charge or discharge.
- It is defined as: \( \tau = \mathrm{R C} \), where \( \mathrm{R} \) is the resistance and \( \mathrm{C} \) is the capacitance.
- For a charging capacitor, \( \tau \) represents the time required for the capacitor’s charge to increase from zero to approximately $63\%$ of its final value.
- For a discharging capacitor, \( \tau \) represents the time required for the capacitor’s charge to decrease from fully charged to approximately $37\%$ of its initial value.
Charging of a Capacitor:
Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, allowing charge to flow easily to its plates.
- As the capacitor charges, the potential difference across the capacitor increases, the current in the circuit branch decreases, and the electric potential energy stored in the capacitor increases.
- The potential difference across the capacitor (\( V_\mathrm{C} \)), the current in the branch (\( I \)), and the stored energy (\( \mathrm{U} \)) all change over time but asymptotically approach steady state after a long time interval.
- After a long time, the charging capacitor reaches its maximum potential difference \( V_\mathrm{max} \), and the current in the branch becomes zero.
Relevant equations for charging:
\( \mathrm{Q}(t) = \mathrm{C} \mathrm{V} \left(1 – e^{-t/\mathrm{RC}}\right) \)
\( V_\mathrm{C}(t) = \mathrm{V} \left(1 – e^{-t/\mathrm{RC}}\right) \)
\( I(t) = \frac{\mathrm{V}}{\mathrm{R}} e^{-t/\mathrm{RC}} \)
Energy stored in the capacitor:
\( \mathrm{U}(t) = \dfrac{1}{2} \mathrm{C} V_\mathrm{C}^2(t) \)
Discharging of a Capacitor:
- Immediately after a charged capacitor begins discharging, the charge on the plates and the energy stored in the capacitor begin to decrease.
- As the capacitor discharges, the charge \( \mathrm{Q} \), potential difference \( V_\mathrm{C} \), and current \( I \) in the branch all decrease over time.
- After a long time (much greater than the time constant), the capacitor is considered fully discharged, and the current in the circuit branch reaches zero.
Relevant equations for discharging:
\( \mathrm{Q}(t) = \mathrm{Q}_0 e^{-t/\mathrm{RC}}\)
\( \mathrm {V}_\mathrm{C}(t) = \mathrm {V}_0 e^{-t/\mathrm{RC}}\)
\( \mathrm {I(t) }= \frac{ \mathrm {V_0}}{\mathrm{R}} e^{-t/\mathrm{RC}} \)
Energy stored in the capacitor:
\( \mathrm{U}(t) = \dfrac{1}{2} \mathrm{C} V_\mathrm{C}^2(t) \)
Steady-State Behavior:
- For times much greater than the time constant (\( t \gg \tau \)), the capacitor and the circuit branch can be modeled using steady-state conditions.
- Charging: Capacitor acts like an open circuit; \( V_\mathrm{C} = V_\mathrm{max} \), \( I = 0 \).
- Discharging: Capacitor acts like a short circuit; \( V_\mathrm{C} = 0 \), \( I = 0 \).
Summary of RC Circuit Behavior:
- The capacitor voltage, current, and stored energy all change continuously over time during charging and discharging.
- The time constant \( \tau \) determines the rate of these changes.
- Immediately after switching, a capacitor behaves as a short (uncharged) or a maximum charge source (fully charged) depending on the context.
- After several time constants (\( t \ge 5 \tau \)), the capacitor reaches steady-state conditions.
Example:
A \(10 \, \mu \mathrm{F}\) capacitor is charged through a \(2 \, k\Omega\) resistor connected to a \(12 \, \mathrm{V}\) battery. Find the time constant, the initial current, and the voltage across the capacitor after 4 ms.
▶️ Answer/Explanation
Time constant: \( \tau = \mathrm{R C} = 2000 \cdot 10 \times 10^{-6} = 0.02 \, \mathrm{s} = 20 \, \mathrm{ms} \)
Initial current: \( I(0) = \frac{\mathrm{V}}{\mathrm{R}} = \frac{12}{2000} = 6 \, \mathrm{mA} \)
Voltage across capacitor after 4 ms: \( V_\mathrm{C}(t) = \mathrm{V} \left(1 – e^{-t/\tau}\right) = 12 \left(1 – e^{-4/20}\right) \approx 2.18 \, \mathrm{V} \)
Example:
A fully charged \(6 \, \mu \mathrm{F}\) capacitor with \( V_0 = 9 \, \mathrm{V} \) is discharged through a \(3 \, k\Omega\) resistor. Find the current after 5 ms and the voltage across the capacitor.
▶️Answer/Explanation
Time constant: \( \tau = \mathrm{R C} = 3000 \cdot 6 \times 10^{-6} = 0.018 \, \mathrm{s} = 18 \, \mathrm{ms} \)
Voltage across capacitor at 5 ms: \( V_\mathrm{C}(t) = V_0 e^{-t/\tau} = 9 e^{-5/18} \approx 6.82 \, \mathrm{V} \)
Current: \( I(t) = \frac{V_\mathrm{C}(t)}{\mathrm{R}} = \frac{6.82}{3000} \approx 2.27 \, \mathrm{mA} \)