2.7 Series circuits
9 exam questions worth 20 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
0 / 20
A 5 Ω resistor and a 10 Ω resistor are connected in series. Calculate the total resistance.
Mark scheme
15 Ω
Model answer
R = 5 + 10 = 15 Ω.
A 12 V battery is connected to a 3 Ω resistor and a 9 Ω resistor in series. Calculate (a) the current, and (b) the potential difference across the 3 Ω resistor.
Mark scheme
total resistance = 12 Ω and I = V / R = 12 / 12 = 1 A (1); V = I R = 1 × 3 (1); 3 V (1)
Model answer
Total resistance = 3 + 9 = 12 Ω. Current = 12 / 12 = 1 A. Potential difference across the 3 Ω resistor = 1 × 3 = 3 V.
Two identical lamps are connected in series to a 6 V battery. State the potential difference across each lamp and explain your answer.
Mark scheme
3 V (1); the supply potential difference is shared equally between identical lamps (1)
Model answer
3 V across each lamp. The potential difference of the supply is shared between the lamps, and as they are identical they share it equally.
A student adds a third lamp in series to a circuit. Explain why the other two lamps become dimmer.
Mark scheme
total resistance of the circuit increases (1); so the current decreases (1)
Model answer
The total resistance of the circuit increases, so the current decreases and the lamps are dimmer.
A string of fairy lights is wired in series. One bulb blows and all the lights go out. Explain why.
Mark scheme
the blown bulb breaks the circuit / the circuit is no longer a complete loop (1); so there is no current in any of the bulbs (1)
Model answer
The blown bulb makes a gap in the single loop, so the circuit is incomplete. There is then no current in any of the bulbs.
Which quantity is the same at every point in a series circuit?
Choose one answer.
Correct Not quite: the answer is Current
Mark scheme
C: current
Model answer
Current (C).
A 12 V battery is connected in series with resistors of 3.0 Ω, 5.0 Ω and 4.0 Ω. Calculate (a) the current in the circuit and (b) the potential difference across the 5.0 Ω resistor.
Mark scheme
total resistance = 12 Ω (1); I = 12 ÷ 12 = 1.0 A (1); V = I R = 1.0 × 5.0 (1); 5.0 V (1)
Model answer
Total resistance = 3.0 + 5.0 + 4.0 = 12 Ω. I = 12 / 12 = 1.0 A. V = 1.0 × 5.0 = 5.0 V.
A 9.0 V battery is connected in series to a 15 Ω resistor and a lamp. The current is 0.20 A. Calculate the resistance of the lamp.
Mark scheme
total resistance = 9.0 ÷ 0.20 = 45 Ω (1); lamp = 45 − 15 (1); 30 Ω (1)
Model answer
Total resistance = 9.0 / 0.20 = 45 Ω. Lamp resistance = 45 − 15 = 30 Ω.
Two 1.5 V cells are connected in series to a lamp. State the total potential difference, and the total if one of the cells is turned round.
Mark scheme
3.0 V (1); 0 V (the p.d.s cancel) (1)
Model answer
3.0 V. If one cell is reversed the two potential differences cancel, so the total is 0 V.
Found a mistake?