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 resistance experiment with an energy check
A fixed resistor is connected to a variable low-voltage supply. The supplied voltage–current data are collected with an ammeter in series and a voltmeter directly across the resistor. The resistor is then held at 6.0 V for 120 s. Treat meter loading as negligible unless considering limitations. Supplied data (Voltage / V; Current / A): (0, 0), (2, 0.1), (4, 0.2), (6, 0.3). (a) Find resistance, charge transferred and electrical energy transferred in the 120 s run. Verify energy using two different expressions. [3 marks] (b) Describe the circuit and evaluate whether a straight line proves the resistor will remain ohmic at all voltages. [3 marks]
Resistor current–voltage data
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Current / A
Horizontal axis (x): Voltage / V
Read the exact data as a table
| Voltage / V | Current / A |
|---|---|
| 0 | 0 |
| 2 | 0.1 |
| 4 | 0.2 |
| 6 | 0.3 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find resistance, charge transferred and electrical energy transferred in the 120 s run. Verify energy using two different expressions.
Reveal the worked solution
- The graph's slope is I/V = 0.050 A V⁻¹, the reciprocal of resistance, so R = 20 Ω.
- At 6.0 V, current is 0.30 A; Q = It = 0.30(120) = 36 C. Voltage is energy transferred per unit charge, giving E = VQ = 216 J.
- A second check is E = I²Rt = (0.30)²(20)(120) = 216 J. Current describes charge flow; it is not consumed by the resistor.
- The ammeter must be in the resistor's current path; the voltmeter connects across its two terminals. Placing an ammeter across the supply can create a low-resistance fault.
- Collect several positive and negative voltages, limit heating and repeat at controlled temperature. A line through the origin supports I∝V over that range.
- At higher power the resistor may warm and change resistance or fail. Real voltmeters draw current and ammeters have resistance; these effects require a loading assessment when comparable to the component values.
Answer (a) R = 20 Ω; Q = 36 C; E = 216 J. (b) The evidence supports constant resistance over the measured range only, under the measurement conditions.
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.
A thermistor alarm with a reversed response
A 6.0 V supply is connected to a 2.0 kΩ fixed resistor in series with an NTC thermistor. Output voltage is measured across the thermistor by a high-resistance input. Its resistance–temperature data are supplied. The alarm activates when output falls below 2.0 V. Supplied data (Temperature / °C; Resistance / kΩ): (10, 8), (20, 4), (30, 2), (40, 1), (50, 0.5). (a) Find output voltage at 20°C and 40°C, and identify the temperature at which the threshold is reached. [3 marks] (b) Predict how the response changes if output is taken across the fixed resistor, and explain why the thermistor is not an ohmic resistor with one universal resistance. [3 marks]
Thermistor calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Resistance / kΩ
Horizontal axis (x): Temperature / °C
Read the exact data as a table
| Temperature / °C | Resistance / kΩ |
|---|---|
| 10 | 8 |
| 20 | 4 |
| 30 | 2 |
| 40 | 1 |
| 50 | 0.5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find output voltage at 20°C and 40°C, and identify the temperature at which the threshold is reached.
Reveal the worked solution
- For a series divider, Vout = 6.0 Rtherm/(2.0 kΩ+Rtherm). At 20°C, this is 6(4)/(2+4) = 4.0 V.
- At 40°C, Vout = 6(1)/(2+1) = 2.0 V. The NTC resistance decreases as temperature rises, so the thermistor voltage also decreases.
- The strict 'below 2.0 V' trigger occurs above 40°C, assuming monotonic response and negligible self-heating. A real alarm needs tolerance and hysteresis specifications.
- The two divider voltages sum to 6.0 V, so the fixed-resistor voltage is 6.0−Vtherm.
- As the thermistor resistance falls, total resistance falls and current rises; the fixed-resistor voltage IRfixed therefore rises.
- Resistance depends strongly on temperature. Even at one bath temperature a large measuring current can cause self-heating, so a calibration requires controlled conditions and suitably low power.
Answer (a) Vout = 4.0 V at 20°C and 2.0 V at 40°C; it activates above 40°C in this ideal calibration. (b) The fixed-resistor output rises with temperature; it is 2.0 V at 20°C and 4.0 V at 40°C.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.