(a) A student claims: 'If the resistance of a conductor is high, no current can flow through it at all.' Critically evaluate this statement using Ohm's law, giving a suitable example to support your answer.
(b) An electric heater coil (R = 50 Ω) and an electric bulb filament (R = 1000 Ω) are connected separately to the same 200 V supply. Calculate the current drawn by each. Which draws more current, and by what factor? Explain why heating appliances are designed to have lower resistance than lighting appliances.
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
(a) The student's claim is incorrect. According to Ohm's law, $V = IR$, which gives $I = V/R$. A high resistance reduces the current but does not make it zero — current will still flow as long as a potential difference is applied. For example, the filament of an electric bulb has a high resistance (~1000 Ω), yet when connected to 220 V, a current of 0.22 A flows through it. Current becomes zero only if the circuit is broken, not merely because resistance is high.
(b)
- Heater: $I = V/R = 200/50 = \mathbf{4\ A}$
- Bulb: $I = V/R = 200/1000 = \mathbf{0.2\ A}$
The heater draws more current, by a factor of 20.
Since power $P = I^2R = V^2/R$, a lower resistance draws more current and dissipates more heat. Heating appliances (irons, heaters) need large amounts of heat energy, so they are designed with low resistance to allow high current and maximum heat production.
Source: Chapter 11 – Electricity, Sections 11.3 (Ohm's Law) and 11.7.1 (Heating Effect Applications)
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Explanation
- Part (a): Examiners want you to explicitly state Ohm's law as $I = V/R$, identify the logical flaw (high R ≠ zero I), and give a concrete numerical or real-life example.
- Part (b): Show both calculations clearly with formula, substitution, and result. State which draws more current and calculate the factor (4 ÷ 0.2 = 20). The final explanation linking $P = V^2/R$ to design logic earns the last conceptual mark.
- Do not write vague statements like "resistance opposes current" without connecting it to Ohm's law mathematically.