Answer to Question #177708 in Electric Circuits for sheherbano haider

Question

A 4 micro Farad capacitor of is to be charged to a potential difference of 100V. Its plates are then disconnected from the source and are connected parallel to other capacitor. The potential difference in this combination comes down to 60V.What is the capacitance of second capacitor.

Answer 1

Explanations & Calculation

• The amount of charge stored in in the first capacitor is

$\qquad\qquad \begin{aligned} \small Q&=\small CV=4\mu F\times 100V \end{aligned}$

• Then what happens when connected in parallel with the other capacitor is that, charges flow between those two until the potential difference between each of their plates become equal to each other, in this case 60V.
• But the sum of the charges stored in both the capacitors are equal to the amount found at the beginning.
• From these relationships we can generate some equations & work this out.

$\qquad\qquad \begin{aligned} \small q_1 +q_2 =\small Q\\ \small C_1V+C_2V&=\small Q\\ \small (C_1+C_2 )V&=\small Q\\ \small (4\mu F+C_2 )\times 60V&=\small 4\mu F\times100V\\ \small 4+C_2&=\small \frac{4\mu F\times100}{60}=6.667\mu\\ \small C_2 &=\small \bold{2.667\mu F} \end{aligned}$

• That is the capacitance of the second capacitor.