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Question

A \( 18 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \,
\text{A} \) to a resistor. What is the resistance of the resistor?

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Explanation

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Terminal voltage: V = ε - I r = 18 - 2 × 3 = 12 V . Resistance: R = (V/I) = (12/2) = 6 Ω . 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

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