O-LevelPhysicsElectricity5054/21 · May/June 2021 · Q78 Marks

(a) Fig. 7.1 shows two circuit components. [Figure 7.1]

(a)For each of these components state what causes an increase in its resistance. thermistor LDR[2]
(b)(i)Fig. 7.2 shows a battery connected to a diode, an ammeter and a resistor. Fig. 7.3 shows the current-voltage graph for the diode. The current in the resistor is 2.0A and the potential difference across the resistor is 5.4 V. State what is meant by the electromotive force (e.m.f.) of the battery.[1]
(b)(ii)Explain why the e.m.f. of the battery must be greater than 5.4V.[2]
(b)(iii)Calculate the electrical energy input to the resistor in 5.0 minutes. energy input =[2]
(b)(iv)The circuit is reconnected with the diode in the opposite direction, as shown in Fig. 7.4. [Figure 7.4] Explain what is observed.[1]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] lower temperature / cooling of thermistor
[B1] less light on LDR
[B1] energy/work per unit charge
[B2] Any 2 from: voltage across diode; voltage across ammeter; voltage across internal resistance (of battery); (e.m.f. / voltage of battery =) 5.4 + 0.6 (or larger)
[C1] (E =) VQ or VIt in any form algebraic or numerical
[A1] 3200 J
[B1] no current/no ammeter reading and diode has been reversed / diode does not conduct/ has (very) high resistance / blocks current / reference to current/voltage graph

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