MHT CET · Physics · Alternating Current
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
- A \(\frac{4}{\sqrt{7}} \mathrm{~A}\)
- B \(1.0 \mathrm{~A}\)
- C \(\frac{4}{7} \mathrm{~A}\)
- D \(0.8 \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(0.8 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
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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