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Question

A battery of emf \( 12 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 5
\, \Omega \) resistor. What is the power dissipated in the external resistor?

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Explanation

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Current: I = (ε/R + r) = (12/5 + 1) = 2 A . Power: P = I² R = (2)² × 5 = 20 W . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 20 W, consistent

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