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An e.m.f. $E=4 \cos (1000 t)$ volt is applied to an LR circuit of inductance $3 \mathrm{mH}$ and resistance $4 \Omega$. The maximum current in the circuit is
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Verified Answer
The correct answer is:
$0.8 \mathrm{~A}$
For LR circuit,
$Z=\sqrt{R^2+(\omega L)^2}$
Comparing equation $E=4 \cos (1000 t)$ with standard equation, $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$, we get $\omega=1000$ units and $E_0=4 \mathrm{~V}$
$\begin{aligned}
\therefore \quad Z & =\sqrt{16+\left(1000 \times 3^2 \times 10^{-6}\right)} \\
& Z=\sqrt{16+9}=5 \Omega \\
\therefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 \mathrm{~A}
\end{aligned}$
$Z=\sqrt{R^2+(\omega L)^2}$
Comparing equation $E=4 \cos (1000 t)$ with standard equation, $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$, we get $\omega=1000$ units and $E_0=4 \mathrm{~V}$
$\begin{aligned}
\therefore \quad Z & =\sqrt{16+\left(1000 \times 3^2 \times 10^{-6}\right)} \\
& Z=\sqrt{16+9}=5 \Omega \\
\therefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 \mathrm{~A}
\end{aligned}$
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