iGCSE Physics (0625) 4.2.3 Electromotive force and potential difference -Exam Style Questions Paper 2 - New Syllabus
Question
B. the electrical work done by a unit charge passing through a component
C. the mechanical work done by a source in moving a unit charge around a complete circuit
D. the mechanical work done by a unit charge passing through a component
▶️ Answer/Explanation
Detailed solution:
Electromotive force ($e.m.f.$) represents the energy supplied by a source, such as a battery or generator, per unit charge.
It is specifically defined as the electrical work done ($W$) by the source to move a unit charge ($Q$) around a complete circuit.
The mathematical relationship is expressed as $E = \frac{W}{Q}$, where $E$ is measured in Volts ($V$).
Option B describes potential difference, which refers to work done across a single component rather than the whole circuit.
Options C and D are incorrect because $e.m.f.$ specifically concerns electrical work rather than mechanical work.
Thus, Option A is the technically accurate definition according to the syllabus standards.
Question
▶️ Answer/Explanation
Detailed solution:
The e.m.f. ($E$) is the work done per unit charge by the source: $E = \frac{W_{total}}{Q} = \frac{120\text{ J}}{50\text{ C}} = 2.4\text{ V}$.
To find the p.d. ($V$) across one lamp, we first determine the charge $Q_{L}$ passing through each lamp.
Since the lamps are identical and in parallel, the total charge $50\text{ C}$ splits equally, so $Q_{L} = \frac{50\text{ C}}{2} = 25\text{ C}$.
The p.d. across one lamp is the work done per unit charge through that lamp: $V = \frac{W_{lamp}}{Q_{L}} = \frac{40\text{ J}}{25\text{ C}} = 1.6\text{ V}$.
In a parallel arrangement, the p.d. across each branch is the same, so $V$ remains $1.6\text{ V}$ for the pair.
Thus, $E = 2.4\text{ V}$ and $V = 1.6\text{ V}$, which corresponds to option B.
▶️ Answer/Explanation
Detailed solution:
Electromotive force (e.m.f.) is the energy provided by a source, such as a battery, per unit charge.
It is mathematically defined by the equation E= Q W , where W is the electrical work done and Q is the charge.
While potential difference (p.d.) also uses V= Q W , it refers specifically to energy transferred per unit charge passing between two points.
The phrase “driving charge round a complete circuit” uniquely identifies the total energy input from the source.
Current is the rate of flow of charge, I= t Q , and power is the rate of energy transfer, P= t ΔE .
Therefore, the description matches the definition of electromotive force, making Option B the correct choice.
