Question:
A circuit containing inductance of \(\frac{1}{6\pi}\ \text{H}\) and a resistance of \(15\ \Omega\) in series. If an AC voltage of \(100\ \text{V}\) and \(60\ \text{Hz}\) is applied to the above circuit, then the current in the circuit and phase difference between voltage and current respectively are
A circuit containing inductance of \(\frac{1}{6\pi}\ \text{H}\) and a resistance of \(15\ \Omega\) in series. If an AC voltage of \(100\ \text{V}\) and \(60\ \text{Hz}\) is applied to the above circuit, then the current in the circuit and phase difference between voltage and current respectively are
Show Hint
For a series \(R-L\) circuit,
\[
X_L=2\pi fL,
\]
\[
Z=\sqrt{R^2+X_L^2},
\]
and
\[
\tan\phi=\frac{X_L}{R}.
\]
Current lags behind voltage by the phase angle \(\phi\).
Updated On: Jun 26, 2026
- \(4\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{5}\right)\)
- \(5.3\ \text{A}\) and \(\tan^{-1}\left(\frac{3}{4}\right)\)
- \(4\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{3}\right)\)
- \(5.3\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{3}\right)\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the given data.
Inductance is \[ L=\frac{1}{6\pi}\ \text{H}. \] Resistance is \[ R=15\ \Omega. \] Applied AC voltage is \[ V=100\ \text{V}. \] Frequency is \[ f=60\ \text{Hz}. \]
Step 2: Find the inductive reactance.
The inductive reactance is \[ X_L=2\pi fL. \] Substituting the values, \[ X_L=2\pi(60)\left(\frac{1}{6\pi}\right). \] \[ X_L=\frac{120\pi}{6\pi}. \] \[ X_L=20\ \Omega. \]
Step 3: Find the impedance of the \(R-L\) circuit.
For a series \(R-L\) circuit, \[ Z=\sqrt{R^2+X_L^2}. \] So, \[ Z=\sqrt{15^2+20^2}. \] \[ Z=\sqrt{225+400}. \] \[ Z=\sqrt{625}. \] \[ Z=25\ \Omega. \]
Step 4: Find the current in the circuit.
Current is \[ I=\frac{V}{Z}. \] \[ I=\frac{100}{25}. \] \[ I=4\ \text{A}. \]
Step 5: Find the phase difference.
For a series \(R-L\) circuit, the phase angle \(\phi\) is given by \[ \tan\phi=\frac{X_L}{R}. \] Therefore, \[ \tan\phi=\frac{20}{15}. \] \[ \tan\phi=\frac{4}{3}. \] Hence, \[ \phi=\tan^{-1}\left(\frac{4}{3}\right). \]
Step 6: Final conclusion.
Therefore, the current and phase difference are respectively \[ \boxed{4\ \text{A}\ \text{and}\ \tan^{-1}\left(\frac{4}{3}\right)} \] Hence, the correct option is \[ \boxed{(3)} \]
Inductance is \[ L=\frac{1}{6\pi}\ \text{H}. \] Resistance is \[ R=15\ \Omega. \] Applied AC voltage is \[ V=100\ \text{V}. \] Frequency is \[ f=60\ \text{Hz}. \]
Step 2: Find the inductive reactance.
The inductive reactance is \[ X_L=2\pi fL. \] Substituting the values, \[ X_L=2\pi(60)\left(\frac{1}{6\pi}\right). \] \[ X_L=\frac{120\pi}{6\pi}. \] \[ X_L=20\ \Omega. \]
Step 3: Find the impedance of the \(R-L\) circuit.
For a series \(R-L\) circuit, \[ Z=\sqrt{R^2+X_L^2}. \] So, \[ Z=\sqrt{15^2+20^2}. \] \[ Z=\sqrt{225+400}. \] \[ Z=\sqrt{625}. \] \[ Z=25\ \Omega. \]
Step 4: Find the current in the circuit.
Current is \[ I=\frac{V}{Z}. \] \[ I=\frac{100}{25}. \] \[ I=4\ \text{A}. \]
Step 5: Find the phase difference.
For a series \(R-L\) circuit, the phase angle \(\phi\) is given by \[ \tan\phi=\frac{X_L}{R}. \] Therefore, \[ \tan\phi=\frac{20}{15}. \] \[ \tan\phi=\frac{4}{3}. \] Hence, \[ \phi=\tan^{-1}\left(\frac{4}{3}\right). \]
Step 6: Final conclusion.
Therefore, the current and phase difference are respectively \[ \boxed{4\ \text{A}\ \text{and}\ \tan^{-1}\left(\frac{4}{3}\right)} \] Hence, the correct option is \[ \boxed{(3)} \]
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