MYP Physics · topic MYP-13.07
Series and parallel 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 thermistor alarm with a reversed response
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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 battery whose terminal voltage depends on the load
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.