Question 1
The electromotive force (e.m.f.) of a battery is 7.5 V .
Define the term electromotive force.
The electromotive force (e.m.f.) of a battery is 7.5 V .
Define the term electromotive force.
(electrical) work done moving a unit charge around a (complete) circuit
work done AND moving a charge (in a circuit)
The unit of the two electrical quantities electromotive force (e.m.f.) and potential difference (p.d.) is the volt (V).
State one other similarity between e.m.f. and p.d.
both relate to energy per unit charge
State one difference between e.m.f. and p.d.
e.m.f. applies to the whole circuit / source or p.d. to one (or more) component or energy conversion to electrical for e.m.f. or from electrical for p.d.
Define potential difference (p.d.).
work done in passing charge through / across a component
work done per unit charge
State the equation which defines electromotive force (e.m.f.) E.
(definition of emf:) E=W / Q
The e.m.f. of a battery is 9.0 V . The battery is in a circuit.
Calculate the work done by the battery when it moves a charge of 30 C around a complete circuit.
270 J
W = EQ OR 9.0 × 30
The unit of the two electrical quantities electromotive force (e.m.f.) and potential difference (p.d.) is the volt (V).
State one other similarity between e.m.f. and p.d.
both relate to energy per unit charge
Define potential difference (p.d.).
work done in passing charge through / across a component
work done per unit charge
What is the unit of potential difference (p.d.)?
ampere
coulomb
coulomb / second
volt
D
Fig. 8.1 shows a circuit.
Fig. 8.1
The potential difference (p.d.) across component Y is measured with a voltmeter.
On Fig. 8.1, draw the symbol for the voltmeter and its connections to the circuit.
voltmeter connected in parallel with component Y
Fig. 7.1 shows two charged metal plates. X marks the position of the centre of the space between the plates.
Fig. 7.1
During a thunderstorm, an electric field is set up between a cloud and the ground. Charges on the cloud and on the ground are shown in Fig. 7.2.
Fig. 7.2
The lightning shown in Fig. 7.2 discharges a current of 28000 A for 0.0012 s .
The lightning transfers 1.2×108 J of energy.
Calculate the potential difference between the base of the cloud and the ground.
potential difference =
[Total: 7]
(p.d. =) 3.6 MV OR 3.6×106 V
E=I t V OR(V=) E / It OR(V=)E/QOR(V=)1.2×108/34