MHT CET · Physics · Alternating Current
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
- A \(0.8 Å\)
- B \(\frac{4}{7} Å\)
- C \(1.0 Å\)
- D \(\frac{4}{\sqrt{7}} Å\)
Answer & Solution
Correct Answer
(A) \(0.8 Å\)
Step-by-step Solution
Detailed explanation
\(E=E_{0} \cos \omega t\)
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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