A student investigates an electrical circuit. The circuit contains:
- a cell,
- a switch,
- a lamp, and
- an ammeter connected in series.
The circuit is shown below:
(a) The student closes the switch and the lamp lights up.
(i) State the direction of current in the circuit.
(1 mark)
(ii) Explain what happens to the charges in the wires when the switch is closed.
(2 marks)
(b) The student measures a current of 0.4 A in the circuit. The potential difference across the lamp is 3.0 V.
(i) Write down the equation that links current, potential difference and resistance.
(1 mark)
(ii) Calculate the resistance of the lamp.
(2 marks)
(c) The student then adds a second identical lamp in series with the first lamp.
(i) State what happens to the brightness of each lamp compared with before.
(1 mark)
(ii) Explain your answer.
(2 marks)
(d) The student now connects the second lamp in parallel instead of series.
(i) Compare the brightness of the lamps in parallel with the brightness when the lamps were in
series.
(1 mark)
(ii) Explain your answer.
(2 marks)
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How did you do? Click to view the Mark Scheme Answers
(a)(i) From positive terminal to negative terminal. (1)
(a)(ii) Circuit is complete so charges (electrons) can move. (1)
Current is the flow of charge around the circuit. (1)
(b)(i) Equation: V = I × R (1)
(b)(ii) R = V ÷ I = 3.0 ÷ 0.4 (1)
Answer: 7.5 Ω (1)
(c)(i) Lamps are dimmer. (1)
(c)(ii) Total resistance increases. (1)
Current in each lamp is smaller. (1)
(d)(i) Lamps are brighter in parallel. (1)
(d)(ii) Each lamp gets full potential difference. (1)
Current through each lamp is greater than in series. (1)
Total: 12 marks
Foundation level exam style questions