WBJEE 2024 Physics & Chemistry Paper

In a series LCR circuit, we need to find the peak voltage across the inductor given the following data:

The rms voltage across the resistor, VR, is 30 V .

The rms voltage across the capacitor, VC, is 90 V .

The applied voltage, V‌applied ‌, is given by: V‌applied ‌=50√2sin‌ωt. This means the peak value of the applied voltage is V‌peak ‌=50√2V.
First, let's determine the rms voltage of the applied voltage. Since rms voltage is the peak voltage divided by √2, we have:
V‌rms, applied ‌=‌

50√2

√2

=50V
In a series LCR circuit, the total rms voltage is the square root of the sum of the squares of the individual voltage drops. Since the total voltage is applied across the resistor, inductor, and capacitor, we have:

V‌total ‌=√VR2+(VL−VC)2
Given:
‌VR=30V
‌VC=90V
‌V‌rms ‌,‌ total ‌=50V
Rearranging the equation, we find:
‌50=√302+(VL−90)2
‌502=302+(VL−90)2
‌2500=900+(VL−90)2
‌2500−900=(VL−90)2
‌1600=(VL−90)2
Taking the square root of both sides:
VL−90=±40
This gives us two possible values for VL :
VL=130V‌ or ‌VL=50V

To find the peak voltage across the inductor, we need to multiply the rms voltage by √2 (since peak voltage V‌peak ‌=V‌rms ‌×√2) :
For VL=130V:V‌peak, ‌L=130×√2=130√2V
For VL=50V:V‌peak, L ‌=50×√2=50√2V
The correct peak voltage across the inductor that matches the options provided is 50√2V. Therefore, the answer is:
Option D: 50√2V