Generated by claude-sonnet-4-6 · 2026-06-15 06:44 · grounding rag
Model Answer
(i) Electric current through a conductor is directly proportional to the potential difference across its terminals, provided temperature remains constant. This is Ohm's Law: $V = IR$.
Circuit diagram to verify Ohm's Law:
```
+–[Battery]–[Rheostat]–[Key]–+
| |
+–[A (Ammeter, series)]–[R]–+
[V (Voltmeter, parallel across R)]
```
(Ammeter connected in series; voltmeter connected in parallel across the resistor R.)
(ii) An ammeter is connected in series in a circuit to measure current. If its resistance were high, it would reduce the current in the circuit, giving an incorrect reading. Hence, an ammeter must have very low (ideally zero) resistance so it does not alter the current being measured.
(iii)
- The slope of V-I graph = $\frac{V}{I}$ = Resistance.
- Graph A (steeper slope → higher resistance) → Series combination, because $R_s = R_1 + R_2$ is greater than individual resistances.
- Graph B (shallower slope → lower resistance) → Parallel combination, because $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$, so $R_p$ is less than any individual resistance.
Source: Chapter 11 – Electricity, Section 11.3 (Circuit Diagram), Ohm's Law, Section 11.6 (Resistance of a System of Resistors)
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Explanation
- (i) Examiners expect the statement of Ohm's Law, its formula, and a neat labelled diagram with ammeter in series and voltmeter in parallel. Don't skip labels.
- (ii) The key logic is: ammeter is in series → high resistance would disturb the circuit current → must be low resistance. One line is enough.
- (iii) Always link slope of V-I graph to resistance. Series resistance > individual resistance (steeper slope); parallel resistance < individual resistance (shallower slope). State the reason clearly — that's where marks lie.