Answer to Question #301399 in Electric Circuits for Diablo

Question

4 microfarad is charged to a potential of 100volts. Another 6 farad of a capacitor is charged to a potential of 300volts. These capacitors are now joined with plates of like charges connected together. Calculate -

1. The Potential Across Each After Joining

2. The Total Electrical Energy After Joining

3. Will The Energy Stored After Joining Be The Same As The Sum Of Energy Stored Before Joining

Answer 1

Explanations & Calculations

1)

• Charges each capacitor had initially,

$\qquad\qquad \begin{aligned} \small q&=\small (4\times10^{-6}).(100)\\ \small Q&=\small (6).(300) \end{aligned}$

• Each plate has the same (+) and (-) charges of $\small q\,\&\, Q$.

• These are connected parallel to each other, so some charge flows from high potential one to lower.
• Then, if some $\small x$ charge flows in between them,

$\qquad\qquad \begin{aligned} \small V_{balanced}=\frac{Q_{balanced}}{C}= \frac{q+x}{4\times10^{-6}}&= \small \frac{Q-x}{6} \\ \small \frac{4\times10^{-4}+x}{4\times10^{-6}} &=\small \frac{1800-x}{6}\\ \small x&\approx\small8.0\times10^{-4}\,C \end{aligned}$

• And now, you know the exchanged charge and can calculate the charge each capacitor having after connecting parallelly: $\small (q+x)\,\&\,(Q-x)$ .
• Then you can calculate the potential each capacitor experiences; $\small V_{balanced}$

2)

• Total electrical energy is $\small \frac{1}{2}C_1V_{balanced}^2+\frac{1}{2}C_2V_{balanced}^2$

3)

• Just calculate the total energy they had before joining, using the same formula; $\small \frac{1}{2}CV^2$ applying to each capacitor individually.
• Then take the difference between the values obtained for before and after cases.