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7.Alternating Current
hard
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 ......$A$
A
$2$
B
$4$
C
$8$
D
$\frac{{20}}{{\sqrt {13} }}$
(AIPMT-2011)
Solution
Here, Resistance, $R=30\, \Omega$
Inductive reactance, $X_{L}=20 \,\Omega$ at $50\, \mathrm{Hz}$
$\because \quad X_{L}=2 \pi v L$
$\therefore \frac{X_{L}}{X_{L}^{\prime}}=\frac{v}{v^{\prime}}$
$X_{L}^{\prime}=\frac{v^{\prime}}{v} \times X_{L}=\left(\frac{100}{50}\right) \times 20\, \Omega=40 \,\Omega$
Impedance, $Z=\sqrt{R^{2}+\left(X_{L}^{\prime}\right)^{2}}=\sqrt{(30)^{2}+(40)^{2}}$
$=50 \,\Omega$
Current in the coil, $I=\frac{V}{Z}=\frac{200\, \mathrm{V}}{50\, \Omega}=4\, \mathrm{A}$
Standard 12
Physics
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