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Question

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 3 \,
\text{A} \) to a resistor. What is the resistance of the resistor?

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Explanation

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Terminal voltage: V = ε - I r = 12 - 3 × 2 = 6 V . Resistance: R = (V/I) = (6/3) = 2 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 2.0 Ω,

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