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Question -

Keeping the source frequencyequal to the resonating frequency of the series┬аLCR┬аcircuit,if the three elements,┬аL,┬аC┬аand┬аR┬аarearranged in parallel, show that the total current in the parallel┬аLCR┬аcircuitis minimum at this frequency. Obtain the current rms value in each branch ofthe circuit for the elements and source specified in Exercise 7.11 for thisfrequency.



Answer -

An inductor (L), acapacitor (C), and a resistor (R) is connected in parallel witheach other in a circuit where,

L┬а=5.0 H

C┬а=80 ╬╝F = 80 ├Ч 10тИТ6┬аF

R┬а=40 ╬й

Potential of the voltagesource,┬аV┬а= 230 V

Impedance (Z) of the givenparallel┬аLCR┬аcircuit is given as:

Where,

╧Й = Angular frequency

Atresonance,

Hence, the magnitude of┬аZ┬аisthe maximum at 50 rad/s. As a result, the total current is minimum.

Rms current flowing throughinductor L is given as:

Rms current flowing throughcapacitor C is given as:

Rms current flowing throughresistor R is given as:

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