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Alternating Current question

2024 · 5 Apr · Shift 1 · Q87
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Alternating Current question

2024 · 5 Apr · Shift 1 · Q87

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20μF20 \mu \mathrm{F}20μF is ‾\underline{\hspace{2cm}}​ V. JEE Main 2024 (Online) 5th April Morning Shift Physics - Alternating Current Question 14 English
Numerical answer
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Correct answer: 50

The question refers to a given series LCR circuit, but the actual circuit diagram / numerical values of the source, resistor, inductor, frequency, or other voltages are not provided in the prompt.

To find the rms voltage across the capacitor in a series LCR circuit, we would normally use:

  1. Capacitive reactance: XC=1ωCX_C = \frac{1}{\omega C}XC​=ωC1​

  2. Circuit current: I=VZI = \frac{V}{Z}I=ZV​ where Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}Z=R2+(XL​−XC​)2​

  3. Voltage across capacitor: VC=IXCV_C = I X_CVC​=IXC​

Without the circuit data, the value cannot be derived uniquely from first principles from the text alone.

However, since the stored correct answer is given as 50, the intended answer is most likely: VC=50 VV_C = 50\text{ V}VC​=50 V

So the integer answer is: 50\boxed{50}50​

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