wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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?


Open in App
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,


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
LC Oscillator
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon