First slide

AC Circuits

Question

A coil has L = 0.04 H and $R = 12 Ω$ . When it is connected to 220 V, 50 Hz supply the current flowing through the coil, in amperes is

Moderate

A

10.7
B

11.7
C

14.7
D

12.7

Solution

Impedance $Z = \sqrt{R^{2} + 4 π^{2} ν^{2} L^{2}}$
$= \sqrt{{(12)}^{2} + 4 \times {(3 .14)}^{2} \times {(50)}^{2} \times {(0 .04)}^{2}}$ = 17.37 A
Now current $i = \frac{V}{Z} = \frac{220}{17 .37} = 12 .7 Ω$

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