Physics - Capacitor Question with Solution | TestHub

(i) At t = 0, capacitor acts as short - circuit

The the reading of voltmeter will be -5 V
At $\mathrm{t}\to \mathrm{\infty },$ capacitor acts as open circuit and no current flows through voltmeter.

If it is ideal voltmeter then its resistance is infinite & therefore it will read +5V
(ii)

The Charge on the capacitor is given by
$\mathrm{q}={\mathrm{q}}_0(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}})$
as q₀ = RC is the max charge stored

For C1=2C=40 $\mu$F, R1=R=25 k${\mathrm{q}}_{\mathrm{x}}=2\mathrm{CV}\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{2\mathrm{CR}}}\right) \mathrm{ }$

The current through resistance of R=25 k is 

$\mathrm{ }\mathrm{x}=\frac{\mathrm{V}}{\mathrm{R}}\mathrm{ }{\mathrm{e}}^{\frac{-\mathrm{t}}{2\mathrm{CR}}}$

For C₂=C=20 $\mu$F, R₂=2R=50 k
${\mathrm{q}}_{\mathrm{y}}=\mathrm{CV}\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{2\mathrm{CR}}}\right)\mathrm{ }$

The current through resistance of 2R=50k is 

$\mathrm{y}=\frac{\mathrm{V}}{2\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{2\mathrm{CR}}}$

Reading of voltmeter ($∆$V) = Voltage across 2C - Voltage across 2R

$$\begin{gathered} ∆\mathrm{V}=\frac{{\mathrm{q}}_{\mathrm{x}}}{2\mathrm{C}}-\mathrm{y}\left(2\mathrm{R}\right) \\ =\mathrm{V}\left[1-2{\mathrm{e}}^{\frac{-\mathrm{t}}{2\mathrm{CR}}}\right]=0 \end{gathered}$$

(iii)

$$\begin{gathered} \mathrm{\tau }=2\mathrm{RC} \\ =2\times 25\times {10}^3\times 40\times {10}^{-6} \\ =1 \sec \end{gathered}$$
Now Current is given by$\mathrm{i}={\mathrm{i}}_0{\mathrm{e}}^{\left(-\frac{\mathrm{t}}{\mathrm{\tau }}\right)\mathrm{ }}$

at t= 1 sec, Current $\mathrm{i}=\frac{{\mathrm{i}}_{\mathrm{o}}}{\mathrm{e}}$

(iv) After long time no current flows since capacitor act as a open circuit and voltmeter acts as an ideal it's resistance is infinite, therefore current does not allow.