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Science (086) — CBSE Class 10 board question

Q1. [2] § 11.6.1 Resistors in Series § 11.6.2 Resistors in Parallel
Use Ohm's law to determine the potential difference across the 3 Ω resistor in the circuit shown in the following diagram when key is closed :
Previously asked in CBSE board exam
2024 31/2/1 Q25
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Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding rag
Model Answer

The circuit has resistors 1 Ω, 2 Ω, and 3 Ω connected in series with a 2 V cell.

Step 1: Total resistance
$$R_s = 1 + 2 + 3 = 6 \text{ Ω}$$

Step 2: Current in circuit (Ohm's law)
$$I = \frac{V}{R_s} = \frac{2 \text{ V}}{6 \text{ Ω}} = \frac{1}{3} \text{ A}$$

Step 3: Potential difference across 3 Ω resistor
$$V_3 = I \times R = \frac{1}{3} \times 3 = \textbf{1 V}$$

Source: Chapter 11, Section 11.6.1 – Resistors in Series

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Explanation
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