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Q.

In a series LCR circuit, a resistor of 300 Ω, a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is _____ × 104 radians s–1.

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answer is 2.

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Detailed Solution

For maximum current in a series LCR circuit, the resonance condition must be satisfied. The resonant angular frequency ω0\omega_0 is given by:

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

Step 1: Given Data

Resistance R=300ΩR = 300 \Omega (not needed for resonance frequency calculation),

Capacitance C=25C = 25 nF =25×109= 25 \times 10^{-9} F,

Inductance L=100L = 100 mH =100×103= 100 \times 10^{-3} H.

Step 2: Calculate Resonant Angular Frequency ω0\omega_0

ω0=1LC\omega_0 = \frac{1}{\sqrt{L C}}

Substituting the values:

ω0=1(100×103)×(25×109)\omega_0 = \frac{1}{\sqrt{(100 \times 10^{-3}) \times (25 \times 10^{-9})}}

 =12500×106 =10650=2×104 rad/s

 

 ω0=2×104 rad/s 

Final Answer:

2

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