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
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
The centrifugal switch is used to disconnect the starting winding of the motor once the motor approaches its normal operating speed i.e 50% – 70% speed.
During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
EXAMIANS