(a) A battery of electromotive force (e.m.f.) 7.8V and internal resistance r is connected to a filament lamp, as shown in Fig. 6.1. [Figure 6.1] A total charge of 750C moves through the battery in a time interval of 1500s. During this time the filament lamp dissipates 5.7 kJ of energy. The e.m.f. of the battery remains constant.

1. (a)(i)Explain, in terms of energy and without a calculation, why the potential difference across the lamp must be less than the e.m.f. of the battery.[1]
2. (a)(ii)1.the current in the circuit[2]
3. (a)(ii)2.the potential difference across the lamp[2]
4. (a)(ii)3.the internal resistance of the battery.[2]
5. (b)(i)Sketch a circuit diagram showing how two of these three resistors may be connected together to give a combined resistance of 30Ω between the terminals shown. Label the values of the resistances on your diagram.[1]
6. (b)(ii)A potential divider circuit is produced by connecting the three resistors to a battery of e.m.f. 9.0V and negligible internal resistance. The potential divider circuit provides an output potential difference V_OUT of 3.6 V. The circuit diagram is shown in Fig. 6.2. On Fig. 6.2, label the resistances of all three resistors and the potential difference V_OUT. [Figure 6.2][2]

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