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An ideal sinusoidal voltage source v(t) = 230√2 sin(2π × 50t) V feeds an ideal inductor L through an ideal SCR with firing angle α = 0°. If L = 100 mH, then the peak of the inductor current, in ampere, is closest to.
- 20.71
- 0
- 10.35
- 7.32
Correct answer: 20.71
Solution
The peak current through the inductor can be calculated using the relationship between the peak voltage and the inductance, considering the firing angle of the SCR. With a firing angle of 0°, the full peak voltage is applied across the inductor, leading to a peak current of approximately 20.71 A.
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