Current electricityAQA A-Level Physics: Topic test

20 questions, 54 marks

AQA A-Level Physics

Current electricity topic test

Total 54 marks

Name

Class

Date

1
A small electroplating bath is connected to a power supply. It carries a steady current of 2.5 A for 20 minutes while the potential difference across it is 4.0 V.
(a)
What charge passes through the bath in 20 minutes?
[1 mark]
A50 C
B3.0×10³ C
C2.1×10⁻³ C
D1.2×10⁴ C
(b)
What is the energy transferred to the bath in 20 minutes?
[1 mark]
A7.5×10² J
B10 J
C2.0×10² J
D1.2×10⁴ J
(c)
Calculate the resistance of the bath. State the current that would flow if the pd were doubled, assuming the resistance stays constant.
[2 marks]
Total for question 1: 4 marks
2
A filament lamp is rated at 12 V, 36 W. It is connected to a 12 V supply of negligible internal resistance and works at its normal brightness.
(a)
What is the current in the lamp?
[1 mark]
A0.33 A
B432 A
C3.0 A
D6.0 A
(b)
The resistance of the lamp at its working temperature is 4.0 Ω. When the lamp is cold its resistance is much smaller. What is the correct explanation?
[1 mark]
AThe ions in the metal vibrate with a greater amplitude when hot, so the electrons collide with them more often.
BThe number of free electrons in the filament decreases as it gets hotter.
CThe cross-sectional area of the filament decreases greatly when it is hot.
DThe electrons move more slowly through a hot filament because the pd falls.
(c)
The lamp is now connected to a 6.0 V supply. Calculate the current that would flow if the resistance remained 4.0 Ω. Explain whether the actual current is greater or smaller than this value.
[2 marks]
Total for question 2: 4 marks
3
A copper cable of length 25 m has a cross-sectional area of 2.5 mm² and is used to connect an appliance to a supply. The resistivity of copper is 1.7×10⁻⁸ Ω m.
(a)
Calculate the resistance of the cable.
[3 marks]
(b)
The appliance draws a current of 13 A. Calculate the power dissipated in the cable. Explain why a cable with a larger cross-sectional area is used for appliances with a higher power rating.
[4 marks]
Total for question 3: 7 marks
4
A battery has an emf of 12.0 V and an internal resistance of 0.80 Ω. It is connected to a network made of a 2.0 Ω resistor in series with a parallel pair of resistors of resistance 6.0 Ω and 12.0 Ω.
(a)
Calculate the current supplied by the battery, the terminal pd, and the current in the 12.0 Ω resistor.
[6 marks]
(b)
A thick wire of negligible resistance is accidentally connected directly across the terminals of the battery. Calculate the current and the power dissipated in the battery. Explain, in terms of energy, why the terminal pd falls almost to zero and why this situation is dangerous.
[6 marks]
Total for question 4: 12 marks
5
A light sensor circuit has a 10 kΩ fixed resistor in series with a light-dependent resistor (LDR) across a 6.0 V supply of negligible internal resistance. The output pd is taken across the LDR. The resistance of the LDR is 500 Ω in bright light and 100 kΩ in darkness.
(a)
What is the output pd in bright light?
[1 mark]
A0.29 V
B5.7 V
C3.0 V
D6.0 V
(b)
What is the output pd in darkness?
[1 mark]
A0.55 V
B6.0 V
C5.5 V
D3.0 V
(c)
The 10 kΩ fixed resistor is replaced by a 1.0 kΩ fixed resistor. Calculate the output pd in bright light and explain how it compares with the original value.
[2 marks]
Total for question 5: 4 marks
6
A battery of constant emf and constant internal resistance is connected in series with a variable resistor and an ideal ammeter. When the variable resistor is set to 4.0 Ω the current is 1.50 A. When it is set to 9.0 Ω the current is 0.75 A.
(a)
What is the internal resistance of the battery?
[1 mark]
A0.50 Ω
B2.0 Ω
C1.5 Ω
D1.0 Ω
(b)
What is the emf of the battery?
[1 mark]
A6.0 V
B7.5 V
C6.8 V
D13.5 V
(c)
When the variable resistor is set to 4.0 Ω, calculate the total power supplied by the battery and the percentage of this power that is dissipated inside the battery.
[2 marks]
Total for question 6: 4 marks
7
A scanner uses a superconducting magnet winding made of niobium-titanium alloy, which has a critical temperature of 9.2 K. The winding carries a current of 150 A and is cooled by liquid helium at 4.2 K. A designer considers replacing it with a copper winding of the same length, 2.0 km, and cross-sectional area 4.0×10⁻⁶ m². The resistivity of copper is 1.7×10⁻⁸ Ω m at room temperature.
(a)
Calculate the resistance of the copper winding at room temperature.
[3 marks]
(b)
Calculate the power that would be dissipated in the copper winding at 150 A. Explain why the superconducting winding does not have this problem.
[4 marks]
Total for question 7: 7 marks
8
A student designs a temperature sensor using a negative temperature coefficient (ntc) thermistor in series with a fixed resistor R across a 6.0 V supply of negligible internal resistance. The output pd is taken across the thermistor. The resistance of the thermistor is 3.0 kΩ at 20 °C and 500 Ω at 80 °C.
(a)
R is chosen so that the output pd is 3.0 V at 20 °C. Calculate R and the output pd at 80 °C. Explain why the output pd changes as the temperature rises.
[6 marks]
(b)
At 80 °C, calculate the total power supplied by the supply and the energy transferred in one hour. Calculate the power dissipated in the thermistor, and explain why this could make the sensor read too high a temperature.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).