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  • The given circuit has two switches <img alt="\mathrm{S}_1 \, \& \mathrm{~S}_2" src="/latex-image/

The given \mathrm{R}, \mathrm{C} circuit has two switches \mathrm{S}_1 \, \& \mathrm{~S}_2. The switch \mathrm{S}_2 is closed till the capacitor \mathrm{C} attain its maximum possible charge. Then S_2 is opened \& \, S_1 is closed simultaneously till the capacitor releases half of its total stored charge q_0 for a time interval  t_1 then S_1 is opened \&\, \mathrm{S}_2 is closed till the capacitor attains a charge (3/4) qo for a time interval t_2. the \left(\mathrm{t}_1 / \mathrm{t}_2\right) is:

                               

Option: 1

\frac{R_1}{\left(R_1+R_2\right)}


Option: 2

\frac{R_2}{\left(R_1+R_2\right)}


Option: 3

\frac{R_1+R_2}{R_1}


Option: 4

\frac{R_1+R_2}{R_2}


Answers (1)

best_answer

When the switch S_2 is closed, the capacitor is charged to a potential V. Now the switch S_2 is opened & the switch S_1 is closed. The instantaneous charge on the capacitor is

             

\Rightarrow q=q_0 e^{-\frac{t}{R_1 c}}

\text { where } \mathrm{q}_0=\mathrm{CV} \text {, putting } \mathrm{t}=\mathrm{t}_1

\text { for } q=q_0 / 2 \text {, we obtain } e^{-\frac{t_1}{R_1 c}}=\frac{1}{2}

\Rightarrow \mathrm{t}_1=\mathrm{R}_1 \mathrm{C} \ln 2

Again the switch \mathrm{S}_1 is opened \& \mathrm{~S}_2 is closed. Therefore the capacitor starts charging from charge \mathrm{q}_0 / 2 to \mathrm{q}_0 the capacitor starts its charging from charge q_0 / 2  to q_0.
Now instantaneous charge on the capacitor is

q^{\prime}=q_0\left[1-e^{-\frac{t}{\left(R_1+R_2\right) C}}\right]+\frac{q_0}{2} e^{-\frac{t}{\left(R_1+R_2\right) C}}=

q_0\left[1-\frac{1}{2} e^{-\frac{t}{\left(R_1+R_2\right) C}}\right]

A t, t=t_2, q^{\prime}=3 q_0 / 4 \Rightarrow \frac{3 q_0}{4} \quad=q_0\left(1-\frac{1}{2} e^{-\frac{t_2}{\left(R_1+R_2\right) C}}\right)

\Rightarrow \ln 2=e^{\frac{t_2}{\left.R_1+R_2\right) C}}

t_2=\left(R_1+R_2\right) C \ln 2

\therefore \text { the required ratio of the times }=\frac{t_1}{t_2}=\frac{R_1 C \ln 2}{\left(R_1+R_2\right) C \ln 2}

\Rightarrow \frac{t_1}{t_2}=\frac{R_1}{\left(R_1+R_2\right)}

Posted by

Ajit Kumar Dubey

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