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.

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

7.17 Inductance of the Inductor, L = 5.0 H

Resistance of the resistor, R = 40 Ω

Capacitance of the capacitor, C = 80 $μ F =$ 80 $\times 10^{- 6}$ F

Potential of the voltage source, V = 230 V

Impedance Z of the given parallel LCR circuit is given as

$\frac{1}{Z}$ = $\sqrt[]{\frac{1}{R^{2}} + {(\frac{1}{ω L} - ω C)}^{2}}$ where $ω$ = angular frequency

At resonance $\frac{1}{ω L} - ω C$ = 0 or $ω^{2}$ = $\frac{1}{L C}$

$ω = \sqrt[]{\frac{1}{L C}}$ = $\frac{1}{\sqrt[]{5 \times 80 \times 10^{- 6}}}$ = 50 rad/s

The magnitude of Z is the maximum at $ω$ = 50 rad/s. As a result, total current is minimum.

rms current flowing through the Inductor L is given as,

$I_{L}$ = $\frac{V}{ω L}$ = $\frac{230}{50 \times 5}$ = 0.92 A

rms current flowing through the Capacitor C is given as,

$I_{L}$ = $\frac{V}{\frac{1}{ω C}}$ &n

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