Ncert Exercises & Solutions

7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C, and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

Given,

rms voltage, 𝑉 = 230 V

Inductance, 𝐿 = 5 H

Resistance, 𝑅 = 40 Ω

Capacitance, 𝐶 = 80 μF

As the circuit elements are arranged in parallel, the impedance of the circuit is
given by,

$\frac{1}{Z^{2}} = \frac{1}{X_{L}^{2}} + \frac{1}{X_{C}^{2}}$

Also, source frequency is equal to the resonance frequency-

$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{5 H \times 80 μ F}} = 50 r a d s^{- 1}$

As the value of 𝑍𝑍 is maximum at this frequency. Hence, the total current will be minimum.

The rms current in the inductor,

I_{L} = \frac{V}{ωL} = \frac{230}{50 \times 5} = 0 .92 A

The rms current in the capacitor,

$I_{L} = V ω C = 230 \times 50 \times 80 \times 10^{- 6} = 0.92 A$

The rms current in the resistor,

$I_{R} = \frac{V}{R} = \frac{230}{40} = 5.75 A$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions