Science (086) — CBSE Class 10 board question

Given: R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 4 Ω, R₄ = 6 Ω, R₅ = 10 Ω, V = 20 V

Step 1: R₃ and R₄ are in series → R₃₄ = 4 + 6 = 10 Ω

Step 2: R₂ (10 Ω) and R₃₄ (10 Ω) are in parallel:
$$\frac{1}{R_{parallel}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \Rightarrow R_{parallel} = 5 \text{ Ω}$$

(i) Effective resistance:
R₁ is in series with R_parallel:
$$R_{eff} = 5 + 5 = \boxed{10 \text{ Ω}}$$

(ii) Current drawn from battery (Ohm's law):
$$I = \frac{V}{R_{eff}} = \frac{20}{10} = \boxed{2 \text{ A}}$$

(iii) Potential difference across R₁ (5 Ω):
$$V_1 = I \times R_1 = 2 \times 5 = \boxed{10 \text{ V}}$$

Source: Chapter 11, Sections 11.6.1 and 11.6.2

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• Circuit reading: Identify series/parallel groupings carefully. R₃ + R₄ are in series (same branch), their combination is parallel to R₂, and the whole parallel block is in series with R₁.
• Examiner looks for: correct identification of combination type, proper formula application, and 3 clearly labelled answers — one for each sub-part.
• Common mistake: Adding all resistors directly or misidentifying which resistors are parallel. Always reduce step by step.
• The voltmeter reads V across R₁ only (it is connected across R₁), so the answer to (iii) is simply I × R₁.