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  4. An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 Ω is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

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Q. An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 $\Omega$ is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

Solution:

Since $\frac{100}{8} =2\pi \times 50 \times L \Rightarrow L= \frac {1}{8\pi}H$
Now $X′_L =2\pi \times 40\times \frac{1}{8\pi}=10\Omega$
$R = 10\Omega$
$Z = \sqrt{10^2+10^2}=10\sqrt{2}\Omega$
$l = \frac{120}{10\sqrt{2}}=6\sqrt{2}A$