Question

A $20F$ capacitor is charged to $5V$ and isolated. It is then connected in parallel to an uncharged capacitor of $30F$. The decrease in the energy of the system will be:

• A. $200J$
• B. $25J$
• C. $150J$
• D. $125J$

Answer 1

The correct option is C

$150J$

Step 1: Given Data

The capacitance of the charged capacitor $C_1=20F$

The voltage of the charged capacitor $V=5V$

The capacitance of the uncharged capacitor $C_2=30F$

Step 2: Calculate Initial Charge

We know that the charge of a capacitor is given as,

$Q_1=C_{1}V$

$\therefore Q_1=20\times 5=100C$

Step 3: Calculate Final Charges

Both the capacitors are connected in parallel.

The total charge of both the capacitors must remain conserved.

$\therefore Q_1+Q_2=100$ $\left(1\right)$

Since the voltage across both the capacitors is the same we can write,

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$

$\Rightarrow Q_1=\frac{C_1}{C_2}{Q}_2$

$\Rightarrow Q_1=\frac{20}{30}{Q}_2$

$\Rightarrow Q_1=\frac{2}{3}{Q}_2$

Upon substituting this in equation $\left(1\right)$ we get,

$\frac{2}{3}{Q}_2+Q_2=100$

$\Rightarrow \frac{5}{3}{Q}_2=100$

$\Rightarrow Q_2=60C$

$\therefore Q_1=100-60=40C$

Step 4: Calculate decrease in energy

The energy of a capacitor is given as,

$U=\frac{1}{2}C{V}^2$ or $U=\frac{1}{2}\frac{Q^2}{C}$

Therefore, the decrease in energy is given by,

$∆U=U_i-U_f$

$\Rightarrow ∆U=\frac{1}{2}{C}_1{V}^2-\left[\frac{1}{2}\left(\frac{{Q_1}^2}{C_1}+\frac{{Q_2}^2}{C_2}\right)\right]$

$\Rightarrow ∆U=\frac{1}{2}\times 20\times 5^2-\left[\frac{1}{2}\left(\frac{{40}^2}{20}+\frac{{60}^2}{30}\right)\right]$

$\Rightarrow ∆U=250-100$

$\Rightarrow ∆U=150J$

Hence, the correct answer is option (C).