Question:

A resistor of \(10\,\Omega\) and an inductor of \(0.1\,\text{H}\) are connected in series to a \(100\ \text{V}, 50\ \text{Hz}\) AC source. The impedance of the circuit is:

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Impedance in RL circuit: \(Z=\sqrt{R^2+X_L^2}\).
Updated On: Jan 5, 2026
  • \(10\,\Omega\)
  • \(\sqrt{100+100\pi^2}\,\Omega\)
  • \(\sqrt{100+10000\pi^2}\,\Omega\)
  • \(100\,\Omega\)
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The Correct Option is B

Solution and Explanation

\[ X_L = 2\pi f L = 2\pi(50)(0.1)=10\pi \] \[ Z=\sqrt{R^2+X_L^2}=\sqrt{10^2+(10\pi)^2}=\sqrt{100+100\pi^2} \]
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