AP Physics · topic 11
Electric Circuits
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A discharging capacitor transfers finite energy to a resistor
A 100 μF capacitor initially at 10.0 V discharges through a 20.0 kΩ resistor. The supplied voltages follow V(t) = V0 exp(−t/RC). There is no power supply during discharge. Use ln2 = 0.693 and neglect leakage and meter loading. Supplied data (Time / s; Voltage / V): (0, 10), (1.386, 5), (2.772, 2.5), (4.158, 1.25). (a) Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value. [3 marks] (b) Find energy remaining and energy dissipated when voltage halves. Derive the energy decay law and explain why current also halves at that time. [3 marks]
Capacitor discharge voltage
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Voltage / V
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Voltage / V |
|---|---|
| 0 | 10 |
| 1.386 | 5 |
| 2.772 | 2.5 |
| 4.158 | 1.25 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value.
Reveal the worked solution
- The time constant is RC = (2.0×10⁴)(1.0×10⁻⁴) = 2.0 s; initial current magnitude is V0/R = 10/20000 = 0.50 mA.
- Initial energy is ½CV0² = ½(1.0×10⁻⁴)(100) = 0.0050 J.
- Setting V/V0 = 1/2 gives t = RC ln2 = 1.386 s, consistent with the supplied data. The time constant is not the voltage half-time.
- Since U = ½CV² and C is fixed, halving voltage quarters stored energy, leaving 5.0/4 = 1.25 mJ.
- The resistor receives the missing 3.75 mJ as thermal energy. Substitution of V(t) gives U(t) = U0 exp(−2t/RC), so energy decays with a time constant RC/2.
- Ohm's law gives I(t) = V(t)/R, so current halves when voltage halves. Integrating I²R over the full discharge returns the finite initial capacitor energy.
Answer (a) RC = 2.0 s; Iinitial = 0.50 mA; Uinitial = 5.0 mJ; voltage half-time = 1.386 s. (b) Remaining energy = 1.25 mJ; dissipated = 3.75 mJ; U = U0 exp(−2t/RC).
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A battery whose terminal voltage depends on the load
A battery supplies an external load. Its measured terminal-voltage versus total-current data are supplied. A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across its terminals. Model the battery as a fixed emf in series with internal resistance. Supplied data (Current / A; Terminal voltage / V): (0, 12), (1, 11), (2, 10), (3, 9). (a) Infer emf and internal resistance, then find total current, terminal voltage and both branch currents. [3 marks] (b) Calculate useful load power and battery internal heating, then predict the load voltage if the 3 Ω branch is disconnected. [3 marks]
Battery load test
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Terminal voltage / V
Horizontal axis (x): Current / A
Read the exact data as a table
| Current / A | Terminal voltage / V |
|---|---|
| 0 | 12 |
| 1 | 11 |
| 2 | 10 |
| 3 | 9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer emf and internal resistance, then find total current, terminal voltage and both branch currents.
Reveal the worked solution
- V = ε−Ir has intercept ε = 12 V and gradient −r = −1.0 V A⁻¹, so internal resistance is 1.0 Ω.
- The parallel load is (1/6+1/3)⁻¹ = 2.0 Ω. Total current is ε/(Rload+r) = 12/3 = 4.0 A, giving terminal voltage 12−4 = 8.0 V.
- Both branches have 8.0 V across them, so I6 = 8/6 = 4/3 A and I3 = 8/3 A. Their sum checks the junction current rule.
- Power delivered externally is VI = 8(4) = 32 W. Internal heating is I²r = 16 W, and εI = 48 W checks energy conservation.
- With only the 6 Ω branch, total current is 12/(6+1) = 12/7 A.
- Terminal voltage is I(6) = 72/7 V, greater than before because the internal voltage drop is smaller. The emf is unchanged in this model.
Answer (a) emf = 12 V; r = 1.0 Ω; total current = 4.0 A; terminal voltage = 8.0 V; branch currents = 4/3 A and 8/3 A. (b) Load power = 32 W; internal heating = 16 W; disconnected-case terminal voltage = 72/7 ≈10.3 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.