SATHEE: Shortcut Methods

Shortcut Methods

JEE Numerical:

1. An LCR circuit has L=1H, C=100μF, and R=10Ω. Calculate the resonant frequency.

$f_{r} = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π (1) (100 \times 10^{- 6})}$

$f_{r} = 159.2 H z$

2. A 100mH inductor, a 10μF capacitor, and a 10Ω resistor are connected in series to a 100V, 50Hz AC source.

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

$Z = \sqrt{10^{2} + (2 π 50 \times 0.1 - 1 / 2 π 50 \times 100 \times 10^{6})^{2}}$

$Z = \sqrt{100 + 2500 - 10} = 44.7 Ω$

$I = \frac{E}{Z} = \frac{100}{44.7} = 2.24 A$

CBSE Numerical:

1. A 50Hz alternating current source of 100V is connected to a series LCR circuit containing R = 100Ω, L = 0.2H, and C = 100μF. Determine the impedance of the circuit.

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

$Z = \sqrt{100^{2} + (2 π 50 \times 0.2 - 1 / 2 π 50 \times 100 \times 10^{6})^{2}}$

$Z = \sqrt{10000 + 100 - 10} = 99.9 Ω$

2. Calculate the resonant frequency of a series LCR circuit having L=10mH, C=400μF, and R=20Ω.

$f_{r} = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π (10 \times 10^{- 3}) (400 \times 10^{- 6})}$

$f_{r} = 79.6 H z$

Shortcut Methods

JEE Numerical:

CBSE Numerical:

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Shortcut Methods

JEE Numerical:

CBSE Numerical:

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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