O-LevelPhysicsElectricity5054/21 · May/June 2024 · Q511 Marks

Fig. 5.1 shows part of a circuit containing a 240V mains supply connected to a lamp and two heaters of resistance 40Ω and 60Ω. [Figure 5.1]

(a)Complete Fig. 5.1 by drawing the symbol for a fuse in the gap in the circuit.[1]
(b)(i)On Fig. 5.2 sketch a current–voltage graph for a filament lamp. [Figure 5.2][1]
(b)(ii)Explain the shape of the current–voltage graph drawn in (i).[2]
(c)Calculate the current at point P in the circuit shown in Fig. 5.1. current = ........................................................... A[3]
(d)(i)The current in the lamp is 1.5A. Suggest a suitable fuse rating for the fuse.[1]
(d)(ii)Explain why a fuse rating much larger than the value in (i) is not suitable.[2]
(d)(iii)Explain why it is necessary to connect a fuse in the live wire rather than the neutral wire or earth wire.[1]

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Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[1] correct symbol;
[1] correct curvature
[1] (at higher current / voltage) temperature increases
[1] (at higher current / voltage) resistance increases
[1] (I =) V/ R in any form
[1] 4 (A) or 6 (A) seen or Rtotal = 24 (Ω)
[1] 10 (A)
[1] any integer between and including 12 and 19 (A)
[1] fuse does not melt / blow or circuit does not shut down aaw
[1] quickly enough or when there is a fault or when the current is large / dangerous
[1] when fuse melts nothing is live / electrocution not possible OR supply of voltage is cut OR voltage is supplied by live wire.

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