A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connec

Q: Why is energy lost when charge is just moving from one capacitor to another?

When charge flows through the connecting wires, it encounters resistance. This flow of current generates heat ($I^2R$ loss), and sometimes even a spark. This dissipated heat accounts for the lost electrical potential energy.

Q: Can I use the formula $\frac{1}{2} \frac{q^2}{C}$ instead of $\frac{1}{2}CV^2$?

Yes, absolutely! For the initial state, $U_i = \frac{q^2}{2C}$. For the final state, the total charge $q$ is conserved, so $U_f = \frac{q^2}{2C_{eq}}$. This is often faster because you don't even need to calculate the common potential!

Q: What happens to the lost energy?

The lost energy is dissipated as heat in the connecting wires and as electromagnetic radiation if a spark occurs.

Q: Is the charge always conserved in such circuits?

Yes, for an isolated system of capacitors, the total net charge remains constant. Charge simply redistributes among the available plates.

Q: Is there a direct formula for energy loss?

Yes, the energy loss when two capacitors are connected is given by $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$. Using this formula can save you a lot of calculation time.

Question

A capacitor $C$ is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is

Options:

(A) $\frac{1}{3}CV_0^2$

(B) $\frac{1}{6}CV_0^2$

(C) $\frac{1}{2}CV_0^2$

(D) $\frac{1}{4}CV_0^2$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Capacitors

Imagine you have a bucket full of water, and another bucket that is completely empty. If you connect them with a pipe, water will flow from the full bucket to the empty one until the water level in both is exactly the same.

This is exactly what happens in our problem! We start with a capacitor of capacitance $C$ that is fully charged to a voltage $V_0$. It holds a certain amount of charge, just like the full bucket.

Then, we disconnect the battery and connect this charged capacitor in parallel with an uncharged capacitor of capacitance $\frac{C}{2}$.

### The Law of Conservation of Charge

When we connect them, charge begins to flow. It flows from the charged capacitor to the uncharged one. But when does it stop? It stops when the electrical pressure, or the **potential**, becomes equal across both capacitors. Let's call this new common potential $V_C$.

Even though the charge is moving around, the **total charge in the system remains strictly conserved**. No charge is created or destroyed.

Let's write this down mathematically. The initial total charge is just the charge on the first capacitor:

$$Q_{initial} = C V_0$$

The final total charge is the sum of the charges on both capacitors, which are now at the common potential $V_C$:

$$Q_{final} = C V_C + \frac{C}{2} V_C = \left(C + \frac{C}{2}\right) V_C = \frac{3C}{2} V_C$$

Equating the initial and final charges, we can easily find the common potential:

$$C V_0 = \frac{3C}{2} V_C \implies V_C = \frac{2}{3} V_0$$

### Calculating the Energy Loss

Now, let's talk about energy. The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.

The initial energy of our system was entirely in the first capacitor:

$$U_i = \frac{1}{2} C V_0^2$$

After the charge redistribution, the final energy is stored in the equivalent parallel combination of the two capacitors, which is $C_{eq} = C + \frac{C}{2} = \frac{3C}{2}$, at the new common potential $V_C$:

$$U_f = \frac{1}{2} C_{eq} V_C^2 = \frac{1}{2} \left(\frac{3C}{2}\right) \left(\frac{2}{3} V_0\right)^2$$

Let's carefully square the term and simplify:

$$U_f = \frac{1}{2} \cdot \frac{3C}{2} \cdot \frac{4}{9} V_0^2 = \frac{1}{3} C V_0^2$$

Finally, the energy loss is simply the difference between the initial and final energies:

$$\Delta U = U_i - U_f = \frac{1}{2} C V_0^2 - \frac{1}{3} C V_0^2 = \frac{1}{6} C V_0^2$$

### The Secret Shortcut

Did you know there is a direct formula for this? Whenever two capacitors $C_1$ and $C_2$ at initial potentials $V_1$ and $V_2$ are connected in parallel, the energy dissipated as heat is given by:

$$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$

If you plug in $C_1 = C$, $V_1 = V_0$, $C_2 = \frac{C}{2}$, and $V_2 = 0$, you will get $\frac{1}{6} C V_0^2$ in just one line! This is a fantastic tool to keep in your arsenal for competitive exams.

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Visualized Solution

$I ni t ia l S t a t e$

$C o n ser v a t i o n o f C ha r g e$

$C o mm o n P o t e n t ia l$

$C a l c u l a t in g V_{C}$

$I ni t ia l E n er g y$

$F ina l E n er g y$

$C a l c u l a t in g F ina l E n er g y$

$E n er g y L oss$

$T h e D i r ec t F or m u l a$

The Sigma Insight: Combination of Capacitors

The Setup

The Law of Conservation of Charge

Calculating the Energy Loss

The Secret Shortcut

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Common Potential

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating VC​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Final Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Loss

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Direct Formula