Given Inductance L = 2 H Rate of change of current di/dt = 5 A/sec Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt) = −(5 × 2) = −10V
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