The net impedance of circuit (as shown in figure) will be : 1. 25 Ω 2. 10√(2) Ω 3. 15 Ω 4. 5√(5) Ω

Answer: 4. 5√(5) Ω For a series LCR circuit Z=√(R²+(X_L-X_C)²) . X_L=2π fL=2π(50)50/π×10⁻³=5 Ω . X_C=1/2π fC=dfrac12π(50)10³/π×10⁻⁶=10 Ω . Z=√(10²+(5-10)²)=√(125)=5sqrt5 Ω . The answer is 5sqrt5 Ω .

Topic: Reactance, impedance, LCR series circuit, resonance· Electromagnetic Induction and Alternating Currents

The net impedance of circuit (as shown in figure) will be :

series circuit: inductor 50/pi mH, capacitor 10^3/pi uF, resistor 10 ohm, source 220 V 50 Hz

1. \(25\,\Omega\)

2. \(10\sqrt{2}\,\Omega\)

3. \(15\,\Omega\)

4. \(5\sqrt{5}\,\Omega\)

Doubtly Asked 30 Oct 2026 NEET UG 2023 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. \(5\sqrt{5}\,\Omega\)

For a series LCR circuit \(Z=\sqrt{R^2+(X_L-X_C)^2}\).

\(X_L=2\pi fL=2\pi(50)\dfrac{50}{\pi}\times10^{-3}=5\,\Omega\).
\(X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\pi(50)\dfrac{10^3}{\pi}\times10^{-6}}=10\,\Omega\).
\(Z=\sqrt{10^2+(5-10)^2}=\sqrt{125}=5\sqrt5\,\Omega\).

The answer is \(5\sqrt5\,\Omega\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

The net impedance of circuit (as shown in figure) will be :

1 answer

Your answer