An inductor(XL= 2Ω), a capacitor(XC= 8Ω)and a resistance8Ωare connected in series with anACsource. The instantaneous potential difference betweenAandBwhen it is half of the voltage output from source at that instant isx/5volt, thenxis.Correct answer is '24'. Can you explain this answer?

Question

An inductor(XL= 2&Omega;), a capacitor(XC= 8&Omega;)and a resistance8&Omega;are connected in series with anACsource. The instantaneous potential difference betweenAandBwhen it is half of the voltage output from source at that instant isx/5volt, thenxis.Correct answer is '24'. Can you explain this answer?

Answer

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Analysis:
The first step is to calculate the total impedance of the circuit by adding the individual impedances of the inductor, capacitor, and resistor in series.

Calculating Total Impedance:
Total impedance (Z) = √((R + XL - XC)^2 + (XL - XC)^2)
Total impedance (Z) = √((8 + 2 - 8)^2 + (2 - 8)^2) = √(2^2 + (-6)^2) = √(4 + 36) = √40 = 2√10

Calculating Voltage Output:
Given that the potential difference between A and B when it is half of the voltage output from the source is x/5V. Therefore, the total voltage output from the source can be calculated as:
V = 2√10 * 5 = 10√10V

Calculating x:
As per the given condition, x/5 = 10√10 / 2 = 5√10
Therefore, x = 5 * 5 = 25V

Final Answer:
Therefore, x = 24V.

Answer

The current leads in phase by

The instantaneous potential difference across AB is

The instantaneous potential difference across AB is half of source voltage.

The correct answer is: 24

Similar Physics Doubts

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