Question

Figure 27-63 shows an ideal battery of emf e= 12V, a resistor of resistance$\mathbf{R}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\Omega }$,and an uncharged capacitor of capacitance $\mathbf{C}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\mu F}$ . After switch S is closed, what is the current through the resistor when the charge on the capacitor $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\mu C}$?

Short answer

The current through the resistor when the charge on the capacitor $8.0\mu C$ is I=2.50A .

Step-by-step solution

Write the given quantities

$$\begin{gathered} e=12V \\ R=4.0\Omega \\ C=4.0\mu F \\ q=8.0\mu C \end{gathered}$$

Determine the concept of capacitance and Ohm’s law  

Voltage across capacitance depends on charge and capacitance. According to Ohm’s law, current is directly proportional to potential difference across a circuit's ends.

Formulae:

Voltage across capacitance C is as follows:

$V_C=\frac{q}{C}$

Current across resistance R is as follows:
role="math" localid="1662365364559" $i_R=\frac{V_R}{R}$

Determine the value of current through the capacitance: 

Voltage across capacitance C is .

$V_c=\frac{8\times {10}^{-6}}{4\times {10}^{-6}}$

Voltage across R is V_R = e - V_c.

V_R= 12-2

=10V

Hence, current through resistance R

$i_R=\frac{V_R}{R}$

Substitute the values and solve as:

$$\begin{gathered} i_R=\frac{10}{4} \\ =2.50A \end{gathered}$$

The current through the resistor when the charge on the capacitor $8.0\mu C$is I=2.50A