Science (086) — AI-generated practice question

Given: V = 15 V, R₁ = 3 Ω, R₂ = 4 Ω, R₃ = 8 Ω (parallel)

(a) Equivalent resistance:

$$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{4} + \frac{1}{8} = \frac{8+6+3}{24} = \frac{17}{24}$$

$$R_p = \frac{24}{17} \approx 1.41 \text{ Ω}$$

(b) Total current:

$$I = \frac{V}{R_p} = \frac{15}{\frac{24}{17}} = \frac{15 \times 17}{24} = \frac{255}{24} \approx 10.6 \text{ A}$$

(c) Current through 8 Ω resistor:

In parallel, voltage across each resistor = 15 V

$$I_3 = \frac{V}{R_3} = \frac{15}{8} = 1.875 \text{ A}$$

Source: Chapter 11, Section 11.6.2

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• In parallel, same voltage (15 V) appears across each resistor — use this directly for part (c).
• For equivalent resistance, use the reciprocal formula: $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.
• Show all steps clearly — CBSE awards marks for the formula, substitution, and final answer separately.
• Keep fractions exact first, then convert to decimals.