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In a $L R$ circuit of $3 \mathrm{mH}$ inductance and $4 \Omega$ resistance, $\operatorname{emf} E=4 \cos 1000 t$ volt is applied. The amplitude of current is
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Verified Answer
The correct answer is:
$0.8 Å$
$E=E_{0} \cos \omega t$
Given $E=4 \cos 1000 t$
From Eqs. (i) and (ii), we get
Peak value of emf, $E_{0}=4 \mathrm{~V}$
Augular fiequency, $\omega=1000 \mathrm{~Hz}$
Now pcak valuc of currcnt is
$$
\begin{aligned}
i_{0} &=\frac{E_{0}}{Z}=\frac{E_{0}}{\sqrt{R^{2}+X_{L}^{2}}} \\
&=\frac{E_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}
\end{aligned}
$$
Putting $E_{0}=4 \mathrm{~V}, R=4 \Omega$,
$$
\begin{array}{l}
\omega=1000 \mathrm{~Hz}, \\
L=3 \mathrm{mH}=3 \times 10^{-3} \mathrm{H}
\end{array}
$$
we get $i_{0}=0.8 \mathrm{~A}$
Given $E=4 \cos 1000 t$
From Eqs. (i) and (ii), we get
Peak value of emf, $E_{0}=4 \mathrm{~V}$
Augular fiequency, $\omega=1000 \mathrm{~Hz}$
Now pcak valuc of currcnt is
$$
\begin{aligned}
i_{0} &=\frac{E_{0}}{Z}=\frac{E_{0}}{\sqrt{R^{2}+X_{L}^{2}}} \\
&=\frac{E_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}
\end{aligned}
$$
Putting $E_{0}=4 \mathrm{~V}, R=4 \Omega$,
$$
\begin{array}{l}
\omega=1000 \mathrm{~Hz}, \\
L=3 \mathrm{mH}=3 \times 10^{-3} \mathrm{H}
\end{array}
$$
we get $i_{0}=0.8 \mathrm{~A}$
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