A student makes a 2.0V battery by connecting two cells of electromotive force (e.m.f.) 2.0V in parallel. The battery, an ammeter with different ranges and three different resistors are used to set up the circuit shown in Fig. 11.1. [Figure 11.1]

1. (a)(i)Explain what is meant by electromotive force.[2]
2. (a)(ii)State one advantage of using two cells in parallel rather than using a single 2.0V cell.[1]
3. (b)(i)Resistor X and the 3.0Ω resistor have a combined resistance that is equal to 2.0Ω. Calculate the total resistance of the circuit,[1]
4. (b)(ii)the resistance of X.[2]
5. (c)(i)Determine the reading of the ammeter.[2]
6. (c)(ii)Suggest a suitable range for the ammeter.[1]
7. (d)The current in the 2.0Ω resistor is I₂. The current in the 3.0Ω resistor is I₃. The current in X is Iₓ. State the equation that relates I₂, I₃ and Iₓ.[1]
8. (e)(i)State the potential difference (p.d.) across the 2.0Ω resistor,[1]
9. (e)(ii)the 3.0Ω resistor.[1]
10. (f)The student sets up a second circuit using a variable d.c. power supply, an ammeter and a 12V metal filament lamp. The circuit is shown in Fig. 11. 2. [Figure 11.2] The d.c. power supply is set to 12V and the ammeter reading is 1.5 A. The student changes the e.m.f. of the d.c. power supply to 6.0V. The lamp dims and the ammeter reading changes.
11. (f)(i)State and explain what happens to the resistance of the metal filament of the lamp.[2]
12. (f)(ii)State whether the new ammeter reading is less than, equal to or greater than 0.75 A.[1]

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