Let the resistance of an electrical device remain constant, while the potential difference across its two ends decreases to one fourth of its initial value. What change will occur in the current through it ? State the law which helps us in solving the above stated question.
Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding rag
Model Answer
By Ohm's law: $I = \dfrac{V}{R}$
Since V decreases to $\dfrac{V}{4}$ and R remains constant:
$$I' = \frac{V/4}{R} = \frac{1}{4} \times \frac{V}{R} = \frac{I}{4}$$
The current will also decrease to one-fourth of its initial value.
Ohm's Law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same. ($V = IR$)
Source: Chapter 11 – Electricity, Section 11.4 Ohm's Law
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Explanation
- The examiner expects two things: (1) the calculated change in current with brief working, and (2) a correct statement of Ohm's Law — both are needed for full 2 marks.
- Always show the substitution step; writing $I' = \frac{V/4}{R}$ demonstrates understanding.
- State Ohm's Law in words (not just the formula) — CBSE mark schemes typically require the verbal statement.