Voltage divider rule:

$\overline{){{\mathbf{V}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{\epsilon}}\frac{{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{R}}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{\epsilon}}\frac{{\mathbf{R}}_{\mathbf{1}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{R}}_{\mathbf{2}}}}$

The charge stored on a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

When the capacitor is fully charged, there is no current through R_{3}. The voltage across the capacitor is equal to the voltage across R_{2}.

V_{C} = V_{2} = ε[R_{2}/(R_{1} + R_{2})] = 42.0[6.00/(6.00 + 8.00)] = 18 V

The capacitor in the figure shown is initially uncharged. The switch is closed at t = 0.

What is the final charge on the capacitor?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the !! Resistor-Capacitor Circuits concept. If you need more !! Resistor-Capacitor Circuits practice, you can also practice !! Resistor-Capacitor Circuits practice problems.