MCQ Solution: Alternating Currents
A coil has resistance 30 ohm and inductive reactance 20 Ohm 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
1) 2.0 A
2) 4.0 A
3) 8.0 A
4) $$\frac{20}{\sqrt{13}}A$$
Solution
$$X_{L}=2\pi fL$$
$$X_{L}\alpha f$$
$$\frac{X_{L_{2}}}{X_{L_{1}}}=\frac{f_{2}}{f_{1}}\Rightarrow X_{L_{2}}=40\Omega $$
$$R=30\Omega$$
$$Z=\sqrt{(30)^{2}+(40)^{2}}=50\Omega $$
$$I=\frac{V}{Z}=\frac{200}{50}=4A$$
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