Question

Obtain the charging time constant of a capacitor in a RC circuit such that current through the resistor is decreased by 50 % of its peak value in 5 seconds.

Answer 1

Answer:0.85

Explanation:

Answer 2

Answer: The value of the time constant is 7.215 seconds.

Explanation: This question is from the RC circuit analysis.

Current in capacitor charged to V in resistor at time t=0 is $I_{0} = V*R$.----(1)

Similarly, current at time t = 5sec is

$I_{5} = V*R*\exp{(-\frac{t}{\tau}) , t=5sec\\ I_{5} = v*R*\exp{(-\frac{5}{\tau})$------------(2) Here $\tau$ is the time constant

Now according to the question,

$\frac{I_{5}}{I_{0}}= \frac{VR\exp{(-5/\tau)}}{VR} = 0.50\\\exp{(-\frac{5}{\tau}) = 0.5$

Take log both sides

$-\frac{5}{\tau} = log0.5\\-\frac{5}{\tau} = -0.693\\\tau = 7.215 sec$

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