Question

A capacitor is fully charged with the help of a battery & then capacitor is removed & connected to same voltage battery but in opposite polarity. Calculate time required to completely discharge the capacitor. (Time constant is 5second ). (1)3.465 s. (2)4.576s (3) 5.687s (4) 6.798sI solve with explanation.​

Answer 1

Answer:

So the capacitor will become discharge after t = 3.465 s

Explanation:

By Kirchoff's law we know that capacitor is initially full charged

$Q = CE$

now for given circuit we have

$E + \frac{(CE - q)}{C} - iR = 0$

so we have

$R \frac{dq}{dt} = \frac{2CE - q}{C}$

now we have

$\frac{dq}{2CE - q} = \frac{dt}{RC}$

now we know that charge on the capacitor will become zero when charge flow through the battery is q = CE

$-ln\frac{2CE - CE}{2CE - 0} = \frac{t}{RC}$

here we know that

$\tau = RC = 5s$

$ln2 = \frac{t}{\tau}$

so we have

$t = 5 ln2$

$t = 3.465 s$

#Learn

Topic : RC Circuit

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