The circuit of Fig. 27-75 shows a capacitor, two ideal batteries, two resistors, and a switch S. Initially S has been open for a long time. If it is then closed for a long time, what is the change in the charge on the capacitor? Assume,$\mathbf{C}\mathbf{=}\mathbf{10}\mathbf{\mu F}$,${\mathbf{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{V}$ ,${\mathbf{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{V}$ , ${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{\Omega }$and${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{\Omega }$.

1. The capacitance of the given capacitor,$C=10\mu F$.
2. Emf of the ideal batteries, ${\epsilon }_1=1.0V$,${\epsilon }_2=3.0V$.
3. Resistance values given are: $R_1=0.20\Omega$,role="math" localid="1662379718034" $R_2=0.40\Omega$.

In the two different switch states, the charge on the capacitor is determined by the emf that is working in the connection process. Thus, when the switch is open, the charge on the capacitor is contributed due to the emf from the battery 2; while, for the condition of the switch closed, the current flow within the circuit is given by the net emf and the total resistance. This further determines the voltage drop across the capacitor and gives the charge to the capacitor. Hence, the difference value of both the open and closed case is the change value of the charge required.

Formulae:

The charge on the capacitor,

q=CV (i)

The voltage equation using Ohm’s law,

V=IR

Here R is the resistance, C is the capacitance of the capacitor and I is the current.

When switch S is open for a long time, the charge on the capacitor is given using equation (i) as follows:

(a)

When switch S is closed for a long time, the current following through both the given resistors can be given using equation (ii) as follows:

$i=\frac{{\epsilon }_2-{\epsilon }_1}{R_1+R_2}$

Substitute the values in the above equation, and we get,

$\begin{array}{l}i=\frac{3.0V-1.0V}{0.20\Omega +0.40\Omega }\\ =3.33A\end{array}$

Now, the voltage drop across the capacitor can be given using equation (ii) as follows:

$V_2={\epsilon }_2-i{R}_2$

Substitute the values in the above equation, and we get,

$$\begin{gathered} V_2=3.0V-\left(3.33A\right)(0.40\Omega ) \\ =1.67V \end{gathered}$$

Now, the above voltage drop occurs for the capacitor from the right loop. Thus, the net change in charge of the capacitor can be given using equations (i) and (a) with the given data as follows:

$\begin{array}{l}q=q_f-q_0\\ =C\left(V_2-{\epsilon }_2\right)\\ \end{array}$

Substitute the values in the above equation, and we get,

$\begin{array}{ll}q=\left(10\times {10}^{-6}+\right)(1.67V& -3.0V)\\ =-13\times {10}^{-6}F& \\ =-13\mu F& \end{array}$

Hence, the required charge value is -13$\mu F$.