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Question

In the series LCR circuit at resonance, the applied ac voltage is 220V. The potential drop across the inductance is 110V. What is the potential drop across the resistance?


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Solution

Step 1: Given Data

The applied ac voltage =220V

The potential drop across the inductance =110V

Series RLC Circuit Analysis & Example Problems | Electrical A2Z

Step 2: Series LCR Circuit at Resonance

We know that the current in a series LCR circuit impressed by an ac emf v=V0cosωt is given by,

I=I0cosωt-ϕ,

Where, I0=V0R2+ωL-1ωC2 and ϕ=tan-1ωL-1ωCR

The RMS current is Ir=VrR2+ωL-1ωC2, 1

Where Vr is the RMS voltage. Thus, Ir depends on the frequency ω of the impressed ac emf. For a certain value of ω, Ir becomes maximum and we say there is resonance.

In this case, ωL=1ωC.

Hence, equation 1 becomes,

Ir=VrR2+02

Ir=VrR 2

Step 3: Potential drop calculation

From the question, it is given that the circuit is at resonance.

When the circuit is at resonance, ωL-1ωC=0, which is why the voltage drop is only utilized by the resistance.

Hence, here we will use equation 2.

The equation 2 is nothing but the form of Ohm's law.

Therefore, Vr=V=220V

Hence, the potential drop across the resistance is 220V,


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