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  • 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.

Q7.17  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.

Answers (1)

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As we know, in the case of a parallel RLC circuit:

 

\frac{1}{Z}=\sqrt{\frac{1}{R^2}+(wC-\frac{1}{wL})^2}

I=\frac{V}{Z}={V}{\sqrt{\frac{1}{R^2}+\left (wC- \frac{1}{wL} \right )^2}}

The current will be minimum when

wC=\frac{1}{wL}

Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency.

RMS value of current in R 

I_{rms}=\frac{V_{rms}}{R}=\frac{230}{40}=5.75A

RMS value in Inductor

I_{inductor}=\frac{V_{rms}}{wL}=\frac{230}{5*50}=0.92A

RMS value in capacitor

I_{capacitor}=\frac{V_{rms}}{1/wL}={230*50*80*10^{-6}}=0.92A

Capacitor current and inductor current will cancel out each other so the current flowing in the circuit is 5.75A.

Posted by

Pankaj Sanodiya

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