1.

A coil has self inductance L = 0.04 H and resistance R = 12 Omega. When it is connected to 220 V, 50 Hz supply, what will be current flowing through the coil ?

Answer»

12.7 A
14.7 A
11.7 A
10.7 A

Solution :`X_L=2pifL=2xx3.14xx50xx0.04= 12.56 Omega`
`|Z|=SQRT(R^2+X_L^2)`
`=sqrt((12)^2 + (12.56)^2)`
`=sqrt(301.75)=17.37 Omega`
`therefore |I|=V/"|Z|"=220/17.37= 12.6655 APPROX` 12.7A


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