Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA No. of turns NA = NB = 1000 Flux linkage in a coil with B = Flux linkage in coil A × 80/100 = 0.8 × 5 × 10−5 = 4 × 10−5 wb =0.04 mwb
Total resistance in the given circuit R = (250 + 250)MΩ = 500 MΩ Current I = V/R = 24/(500 × 103) Now the Voltage in the voltmeter = \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3} V = 12 V
EXAMIANS