SSC JE Electrical 2019 with solution SET-1
Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:
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
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
From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below.
Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R
Here
Current I = 1/2 A = 0.5 A
Voltage V = 5 V
R = V/I = 0.5/5
R = 10Ω
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The average value of the sine wave over one complete cycle is actually zero. Hence, for a sine wave, the average value is defined over half the period. The average value expressed in terms of peak value is given by Average value = 0.637 × peak value
EXAMIANS
