Question

When the uncharged capacitor is charged through a resistor R by a battery of emf E the time taken to get charge that is 50% of the equilibrium value would be
1. (ln2)RC
2.RC/ln2
3.ln2/RC
4.RC/2

Answer 1

Answer:

.RC/ln2 hsiwkehdyeheeh

Answer 2

The time taken to get the charge that is 50% of the equilibrium value would be RCln2.

When the uncharged capacitor is charged through a resistor R by a battery of emf ξ.

We have to find the time taken to get the charge 50% of the equilibrium value.

Potential of a capacitor at a time t, is given by,

$Q=\mathcal{E}C\left(1-e^{-\frac{t}{RC}}\right)$

Where V is the potential difference of capacitor when it is charging after time t, ξ is emf of battery, R is resistance and C is capacitance of capacitor.

Here, charge 50% means,

$Q=\frac{\mathcal{E}C}{2}$

$\implies\frac{\mathcal{E}C}{2}=\mathcal EC\left(1-e^{-\frac{t}{RC}}\right)\\\\\implies\frac{1}{2}=1-e^{-\frac{t}{RC}}\\\\\implies e^{-\frac{t}{RC}}=\frac{1}{2}\\\\\implies\frac{t}{RC}=\text{ln2}\\\\\implies t=\text{RCln2}$

Therefore the time taken to get the charge that is 50% of the equilibrium value would be RCln2.

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