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Standard XII

Physics

More about Q=CV

The given R-C...

Question

The given R-C circuit has two switches $S_{1}$ and $S_{2}$ . Switch $S_{2}$ is closed, and $S_{1}$ is open till the capacitor is fully charged to $q_{0}$ . Then $S_{2}$ is opened and $S_{1}$ is closed simultaneously till the charge on capacitor remains $q_{0} / 2$ for which it takes time $t_{1}$ . Now $S_{1}$ is again opened, and $S_{2}$ is closed till charge on capacitor becomes $3 q_{0} / 4$ . It takes time $t_{2}$ (see Figure for reference). Find the ratio $t_{1} / t_{2}$ .

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Solution

$q_{0} = C E, q_{0} / 2 = q_{0}^{e^{- t 1 / R C}}$ or $t_{1} = R C ln 2$
Now
$3 q_{0} / 4 \int q_{0} / 2 \frac{d q}{C E - q} =_{2} \frac{d t}{3 R C}$
or $t_{2} = 3 R C ln 2$
From here, we get $t_{1} / t_{2} = 1 / 3$ .

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q0=CE,q0/2=qe−t1/RC0 or t1=RCln2Now 3q0/4∫q0/2dqCE−q=t2∫0dt3RC or t2=3RCln2From here, we get t1/t2=1/3.

$q_{0} = C E, q_{0} / 2 = q_{0}^{e^{- t 1 / R C}}$ or $t_{1} = R C ln 2$
Now
$3 q_{0} / 4 \int q_{0} / 2 \frac{d q}{C E - q} =_{2} \frac{d t}{3 R C}$
or $t_{2} = 3 R C ln 2$
From here, we get $t_{1} / t_{2} = 1 / 3$ .