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Physics

Resistivity

For a circuit...

Question

For a circuit shown in Figure switch $S_{1}$ , is closed at $t = 0$ , then at $t = (2 R_{2} + R_{1}) C, S_{1}$ is opened and $S_{2}$ is closed.
a. Find the charge on capacitor at $t = (2 R_{2} + 2 R_{1}) C$ .
b. Find current through $R_{2}$ (adjacent to the battery) at $t = (3 R_{1} + 2 R_{2}) C$ .

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Solution

a. For $t = 0$ to $t = (2 R_{2} + R_{1}) C$ , capacitor gets charged from all the three resistors. So
$q = C E [1 - e^{\frac{1}{(2 R_{2} + R_{1}) C}}]$ .
Putting $t = (2 R_{2} + R_{1}) C$ , we get
$q_{1} = C E [1 - e^{- 1}] = \frac{C E (e - 1)}{e}$ .
Now battery is disconnected and the capacitor gets discharged through $R_{1}$ after $S_{1}$ is opened and $S_{2}$ is closed. So

$\int_{q_{1}}^{q_{2}} \frac{d q}{q} - (2 R_{2} + 2 R_{1}) C \int (2 R_{2} + R_{1}) C \frac{d t}{R_{1} C}$
or $q_{2} = \frac{q_{1}}{e} = \frac{C E (e - 1)}{e^{2}}$

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For a circuit shown in Figure switch S1, is closed at t=0, then at t=(2R2+R1)C,S1 is opened and S2 is closed.a. Find the charge on capacitor at t=(2R2+2R1)C.b. Find current through R2 (adjacent to the battery) at t=(3R1+2R2)C.

For a circuit shown in Figure switch $S_{1}$ , is closed at $t = 0$ , then at $t = (2 R_{2} + R_{1}) C, S_{1}$ is opened and $S_{2}$ is closed.
a. Find the charge on capacitor at $t = (2 R_{2} + 2 R_{1}) C$ .
b. Find current through $R_{2}$ (adjacent to the battery) at $t = (3 R_{1} + 2 R_{2}) C$ .