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Question

In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, εrms = 50 V and ν = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

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Solution

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10−6 F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, εrms = 50 V
Frequency of source, f = 50/π Hz
Reactance of the inductor (XL) is given by,
XL= ωL = 2πfL
XL = 2×π×50π×1 = 100 Ω
Reactance of the capacitance XC is given by,
XC=1ωC = 12πfC
XC = 12π×50π×20×10-6
XC = 500 Ω
(a) Impedance of an LCR circuit Z is given by,
Z=R2+XC-XL2 Z =3002+500-1002 Z =3002+4002 Z =500
RMS value of current Irms is given by,
Irms=εrmsZ
Irms = 50500
Irms = 0.1 A
(b) Potential across the capacitor VC is given by,
VC = Irms × XC
VC = 0.1 × 500 = 50 V
Potential difference across the resistor VR is given by,
VR = Irms × R
VR = 0.1 × 300 = 30 V
Potential difference across the inductor VL is given by,
VL = Irms × XL
VL= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

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