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Question
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 for this frequency.
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Solution
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H
C = 80 μF = 80 × 10−6 F
R = 40 Ω
Potential of the voltage source, V = 230 V
Impedance (Z) of the given parallel LCR circuit is given as:
`1/"Z" = sqrt(1/"R"^2 + (1/(ω"L") - ω"C")^2)`
Where,
ω = Angular frequency
At resonance, `1/(ω"L") - ω"C"` = 0
∴ ω = `1/sqrt("LC")`
= `1/(sqrt(5 xx 80 xx 10^-6))`
= 50 rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum.
Rms current flowing through inductor L is given as:
IL = `"V"/(ω"L")`
= `230/(50 xx 5)`
= 0.92 A
Rms current flowing through capacitor C is given as:
IC = `"V"/(1/(ω"C")) = ω"CV"`
= 50 × 80 × 10−6 × 230
= 0.92 A
Rms current flowing through resistor R is given as:
IR = `"V"/"R"`
= `230/40`
= 5.75 A
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