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Q. If emf $E=4 \cos 1000\, t$ volt is applied to an $L-R$ circuit of inductance $3 \,mH$ and resistance $4\, \Omega$, the amplitude of current in the circuit is

AIIMSAIIMS 2011

Solution:

Impedence, $Z=\sqrt{R^{2}+X_{L}^{2}}$
Here, $R=4 \Omega, X_{L}=L \omega$
$=3 \times 10^{-3} \times 1000\, \Omega$
Then, $=3\, \Omega$
$Z=\sqrt{(4)^{2}+(3)^{2}}$
or $Z=\sqrt{16+9}=\sqrt{25}$
or $Z=5\, \Omega$
Hence, current, $I_{0} =\frac{E_{0}}{Z}$
$=\frac{4}{5}=0.8\, A$