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Q. In the circuit shown in figure, the a.c. source gives a voltage V = 20 cos (2000t) . Neglecting source resistance, the voltmeter and ammeter readings will be
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KCETKCET 2003Electromagnetic Induction

Solution:

Z = $\sqrt{R^2 + ( X_L - X_C)^2}$
Here $R = 6 + 4 = 10\Omega$, $\omega = 2000\, rd\, s^{-1}X_L = \omega\,L = 2000 \times 5 \times 10^{-3} = 10 \Omega$ and
$X_C = \frac{1}{\omega_C}=\frac{1}{200\times50\times10^{-6}}=10\Omega$
$Z=\sqrt {10^2+(10-10)^2} = 10 \Omega$
Current , $I_o=\frac{V_o}{Z}=\frac{20}{10}=2A \therefore I_{rms}= \frac{2}{\sqrt{2}}=\sqrt{2}A$
Voltage $V=\sqrt{2}\times\sqrt{4^2+0}+2\sqrt{2}V$