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# A coil has resistances 30$\Omega$ and inductive reactance $20\Omega$ at 50Hz frequency. If an ac source of 200V, 100Hz is connected across the coil, the current in the coil will be,(A) 13A(B)2.0A(C )4.0A(D)8.0A

Last updated date: 13th Jun 2024
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Hint: First calculate the inductive reactance for 50Hz and 100Hz. And then comparing those two equations we get the inductive reactance for 100Hz. After that calculate the impedance of the circuit. Then finally calculate the value of current using the ac source voltage and impedance.

Formula used:
Inductive reactance, ${{X}_{L}}=\omega L=2\pi fL$ …………….(1)
Impedance, $z=\sqrt{\left( X_{L}^{'2} \right)+\left( {{R}^{2}} \right)}$ ………….(2)
Current, I = $\dfrac{V}{z}$ ……………(3)

Given that,
Resistance, R=30$\Omega$
Inductive reactance, ${{X}_{L}}=20\Omega$
Frequency, f= 50Hz
Using equation (1),
${{X}_{L}}=\omega L=2\pi fL$
Substituting the values of inductive reactance and frequency we get,
$20=2\pi \times 50\times L$ …………(4)
When the frequency of the ac source is changed to 100Hz,
New inductive reactance , $X_{L}^{'}=\omega 'L=2\pi \times 100\times L$
This can be rearranged as,
$X_{L}^{'}=2\pi \times (50\times 2)\times L$ ……….(5)
Substituting equation (4) in (5),
$X_{L}^{'}=2\times {{X}_{L}}=2\times 20=40\Omega$
Substituting $X_{L}^{'}\And R$ in equation (2),
$\Rightarrow$ Impedence, $z=\sqrt{{{40}^{2}}+{{30}^{2}}}=50\Omega$
$\Rightarrow$ Current, $I=\dfrac{V}{z}=\dfrac{200}{50}=4A$

Hence here option(C) is correct.