Question:
The impedance of an LR circuit with
\[
L=\frac{60}{\pi}\ \text{mH},\quad R=8\ \Omega
\]
and frequency \(50\ \text{Hz}\) is
The impedance of an LR circuit with
\[
L=\frac{60}{\pi}\ \text{mH},\quad R=8\ \Omega
\]
and frequency \(50\ \text{Hz}\) is
Show Hint
For an LR circuit:
\[
X_L=2\pi fL
\]
and
\[
Z=\sqrt{R^2+X_L^2}
\]
Always calculate inductive reactance first before finding impedance.
Updated On: Jun 25, 2026
- \(1.3\ \Omega\)
- \(14.3\ \Omega\)
- \(20\ \Omega\)
- \(10\ \Omega\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Convert inductance into henry.
Given, \[ L=\frac{60}{\pi}\ \text{mH} \] Since \[ 1\ \text{mH}=10^{-3}\ \text{H}, \] we get \[ L=\frac{60}{\pi}\times 10^{-3}\ \text{H} \] \[ L=\frac{0.06}{\pi}\ \text{H} \]
Step 2: Calculate inductive reactance.
Inductive reactance is \[ X_L=2\pi fL \] Substituting values, \[ X_L=2\pi(50)\left(\frac{0.06}{\pi}\right) \] \[ X_L=100\times 0.06 \] \[ X_L=6\ \Omega \]
Step 3: Calculate impedance of LR circuit.
Impedance is \[ Z=\sqrt{R^2+X_L^2} \] Substituting \[ R=8\ \Omega,\qquad X_L=6\ \Omega, \] we get \[ Z=\sqrt{8^2+6^2} \] \[ =\sqrt{64+36} \] \[ =\sqrt{100} \] \[ Z=10\ \Omega \]
Step 4: Final conclusion.
Hence, the impedance of the circuit is \[ \boxed{10\ \Omega} \]
Given, \[ L=\frac{60}{\pi}\ \text{mH} \] Since \[ 1\ \text{mH}=10^{-3}\ \text{H}, \] we get \[ L=\frac{60}{\pi}\times 10^{-3}\ \text{H} \] \[ L=\frac{0.06}{\pi}\ \text{H} \]
Step 2: Calculate inductive reactance.
Inductive reactance is \[ X_L=2\pi fL \] Substituting values, \[ X_L=2\pi(50)\left(\frac{0.06}{\pi}\right) \] \[ X_L=100\times 0.06 \] \[ X_L=6\ \Omega \]
Step 3: Calculate impedance of LR circuit.
Impedance is \[ Z=\sqrt{R^2+X_L^2} \] Substituting \[ R=8\ \Omega,\qquad X_L=6\ \Omega, \] we get \[ Z=\sqrt{8^2+6^2} \] \[ =\sqrt{64+36} \] \[ =\sqrt{100} \] \[ Z=10\ \Omega \]
Step 4: Final conclusion.
Hence, the impedance of the circuit is \[ \boxed{10\ \Omega} \]
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