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Question
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 µF, R = 40 Ω.
- Determine the source frequency which drives the circuit in resonance.
- Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
- Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
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Solution
Inductance of the inductor, L = 5.0 H
Capacitance of the capacitor, C = 80 μF = 80 × 10−6 F
Resistance of the resistor, R = 40 Ω
Potential of the variable voltage source, V = 230 V
(a) Resonance angular frequency is given as:
`ω_"R" = 1/sqrt"LC"`
= `1/sqrt (5 xx 80 xx 10^-6)`
= `10^3/20`
= 50 rad s−1
Hence, the circuit will come in resonance for a source frequency of 50 rad s−1.
(b) Impedance of the circuit is given by the relation,
`"Z" = sqrt("R"^2 + (ω"L" - 1/(ω"C"))^2`
At resonance,
`ω"L" = 1/(ω"C")`
∴ Z = R = 40 Ω
Amplitude of the current at the resonating frequency is given as:
`"I"_0 = "V"_0/"Z"`
Where,
V0 = Peak voltage
= `sqrt2 "V"`
∴ `"I"_0 = (sqrt(2) "V")/"Z"`
= `(sqrt2 xx 230)/4`
= 8.13 A
Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) The rms potential drop across the inductor,
`("V"_"L")_"rms" = "I" xx ω_"R""L"`
Where,
I = rms current
= `"I"_0/sqrt2`
= `(sqrt2 "V")/(sqrt2 "Z")`
= `230/40 "A"`
∴ `("V"_"L")_"rms"= 230/40 xx 50 xx 5`
= 1437.5 V
Potential drop across the capacitor,
`("V"_"c")_"rms" = "I" xx 1/(ω_"R" "C")`
= `230/40 xx 1/(50 xx 80 xx 10^-6)`
= 1437.5 V
Potential drop across the resistor,
`("V"_"R")_"rms" = "IR"`
= `230/40 xx 40`
= 230 V
Potential drop across the LC combination,
`"V"_"LC" = "I" (ω_"R" "L" - 1/(ω_"R" "C"))`
At resonance, ωRL = `1/(ω_"R""C")`
∴ VLC = 0
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
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