13. A coil has resistance $30 \Omega$ and inductive reactance $20 \Omega$ at $50 Hz$ frequency. If an AC source, of $200 volt$, $100 Hz$, is connected across the coil, the current in the coil will be: (2011 Mains)
Solution:
Inductive reactance is $X_L = 2\pi f L \propto f$. At $100 Hz$, $X_L' = 20 \times \frac{100}{50} = 40 \Omega$. Circuit impedance is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{30^2 + 40^2} = 50 \Omega$. Current in the coil is $I = \frac{V}{Z} = \frac{200}{50} = 4.0 A$.
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