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Question: A coil has L = 0.04 H and \(R = 12\Omega\). When it is connected to 220V, 50Hz supply the current fl...

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

A

10.7

B

11.7

C

14.7

D

12.7

Answer

12.7

Explanation

Solution

Impedance Z=R2+4π2ν2L2Z = \sqrt{R^{2} + 4\pi^{2}\nu^{2}L^{2}}

=(12)2+4×(3.14)2×(50)2×(0.04)= \sqrt{(12)^{2} + 4 \times (3.14)^{2} \times (50)^{2} \times (0.04)}= 17.37 A

Now current i=VZ=22017.37=12.7Ωi = \frac{V}{Z} = \frac{220}{17.37} = 12.7\Omega