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Determine the peak current if an inductor of inductance 500 mH is connected to an ac source of peak emf 650 V and frequency 100 Hz(a) 1.55 A(b) 2.07 A(c) 7.89 A(d) 9.87 AThis question was addressed to me during an online exam.This key question is from AC Voltage Applied to an Inductor in division Alternating Current of Physics – Class 12

Answer»

Right OPTION is (b) 2.07 A

For explanation I would say: PEAK current (I0) = \(\frac {E_0}{X_L}\)

I0 = \(\frac {650}{(2 \TIMES 3.14 \times 100 \times 0.5)}\)

I0 = 2.07 A

Therefore, the peak current is calculated as 2.07 A.



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