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Physics
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
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Q. A coil has $L=0.04\, H$ and $R=12\, \Omega$. When it is connected to $220 \,V , 50 \,Hz$ supply, the current flowing through the coil, in amperes, is
Alternating Current
A
10.7
B
11.7
C
14.7
D
12.7
Solution:
Impedance, $Z=\sqrt{R^{2}+4 \pi^{2} v^{2} L^{2}}$
$=\sqrt{(12)^{2}+4 \times(3.14)^{2} \times(50)^{2} \times(0.04)}=17.37 A$
Now, current, $i=\frac{V}{Z}$
$=\frac{220}{17.37}=12.7 \,\Omega$