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# A coil has resistance 30 ohm and inductive reactance 20 ohm at 50Hz frequency. If an AC source of 200 volts, 100Hz is connected across the coil, the current in the coilA. 2.0AB. 4.0AC. 8.0AD. 20A

Last updated date: 17th Sep 2024
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Hint: Inductive reactance is directly proportional to frequency of AC source. As frequency increases inductive reactance also increases.

Formula used:
Impedance of the coil,$Z = \sqrt {\left( {{R^2} + {X^2}_L} \right)}$
Where, r is the resistance of the coil (ohm), ${X_L}$ is the inductive reactance (ohm)
Current in the coil,$I = \dfrac{{{V_{rms}}}}{Z}$
${V_{rms}}$ Is the RMS value of voltage (V)

Given, resistance of the coil r=30ohm
The inductive reactance of the coil ${X_L}$=20 ohm
Frequency, f=50hz, ${V_{rms}}$=200v
We know that ${X_L} = \omega L$
$= 2\pi fL$
Then,$20 = \left( {2\pi } \right)50 \times L$ ……………………. (1)
When frequency of ac source is changed to 100hz, then
${X^I}_L = {\omega ^I}L$
$= 2\pi {f^I}L$
Then,${X^I}_L = \left( {2\pi } \right)100L$
Above equation can be rearranged as,
${X^I}_L = \left( {2\pi } \right)50 \times L \times 2$
Now substitute equation (1) in the above equation we get,
${X^I}_L = 20 \times 2 = 40\Omega$
Therefore,
Impedance of the coil,$Z = \sqrt {\left( {{R^2} + {X^I}{{^2}_L}} \right)}$
$Z = \sqrt {\left( {{{30}^2} + {{40}^2}} \right)} = 50\Omega$
Then, current in the coil,$I = \dfrac{{{V_{rms}}}}{Z}$
$I = \dfrac{{200}}{{50}}$
$= 4A$

So, the correct answer is “Option B”.