Three resistors of resistances R₁, R₂ and R₃ are connected as shown in [Figure 6.1]. The total current in the combination of resistors is I and the potential difference across the combination is V. A battery of electromotive force (e.m.f.) 6.0 V and internal resistance r is connected to a resistor of resistance 12 Ω and a variable resistor X, as shown in [Figure 6.2].

1. (a)Show that the total resistance R of the combination is given by the equation 1/R = 1/R₁ + 1/R₂ + 1/R₃[2]
2. (b)(i)By considering energy, explain why the potential difference across the battery's terminals is less than the e.m.f. of the battery.[2]
3. (b)(ii)1.A charge of 2.5 kC passes through the battery. Calculate the total energy transformed by the battery,[2]
4. (b)(ii)2.the number of electrons that pass through the battery.[1]
5. (b)(iii)The combined resistance of the two resistors connected in parallel is 4.8 Ω. Calculate the resistance of X.[1]
6. (b)(iv)Use your answer in (b)(iii) to determine the ratio power dissipated in X / power dissipated in 12 Ω resistor[2]
7. (b)(v)The resistance of X is now decreased. Explain why the power produced by the battery is increased.[1]

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