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A coil of inductance 0.50 H and resistance 100 omega is connected to a 240 V, 50 Hz ac supply.

Q 7.13 (a)   A coil of inductance 0.50Hand resistance 100\Omega is connected to a  240V50Hz  ac supply. What is the maximum current in the coil?

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Given,

The inductance of the coil L=0.50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequencyf=50Hz

Now, as we know peak voltage = \sqrt2(RMS Voltage)

Peak voltage 

V_{peak}=\sqrt2*240=339.4V 

The impedance of the circuit :

Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi (.5) *50)^2}

Now peak current in the circuit :

I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi (.5) *50)^2}}=1.82A

Hence peak current is 1.82A in the circuit.

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