Question 1 Report
A rechargeable lithium-ion cell does 540 J of electrical work in moving 100 C of charge around a complete circuit. A student wants to determine the electromotive force of the cell using the definition that relates e.m.f. to work done and charge. The cell is connected to a fixed resistor of 5.4 Ω in a simple series circuit. Fig. 5.1 shows the circuit. The cell has negligible internal resistance and the connecting wires have negligible resistance.
(a) State the equation that defines electromotive force in terms of work done and charge. [1]
(b) Calculate the e.m.f. of the cell. [2]
(c) Calculate the current flowing through the resistor. [2]
(d) State the unit of e.m.f. [1]
Marking Scheme
Explanation
(a) Electromotive force (e.m.f.) is defined as the work done by a source of electrical energy per unit charge that passes through it:
\(\text{e.m.f.} = \frac{W}{Q}\)
where \(W\) is the work done (energy transferred) in joules and \(Q\) is the charge in coulombs. Although it has "force" in its name, e.m.f. is actually measured in volts and represents an energy-per-charge quantity, not a force.
(b) Substituting the given values:
\(\text{e.m.f.} = \frac{540 \text{ J}}{100 \text{ C}} = 5.4\) V
This means the cell does 5.4 J of work on every coulomb of charge that passes through it.
(c) Since the cell has negligible internal resistance, the terminal voltage equals the e.m.f. (5.4 V). All this voltage appears across the external resistor. Applying Ohm's law:
\(I = \frac{V}{R} = \frac{5.4}{5.4} = 1.0\) A
The numerical coincidence that \(V = R\) gives \(I = 1.0\) A is simply because the resistance was chosen to equal the e.m.f. value.
(d) The unit of e.m.f. is the volt (V), the same as for potential difference. One volt equals one joule per coulomb (1 V = 1 J/C).
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