A 5 micro F capacitor is charged fully by a 220V supply. It is then disconnected from the supply and

Q: Why is energy lost when we connect two capacitors? Where does it go?

When charges flow from one capacitor to another, they travel through connecting wires. Even if the wires have very low resistance, the transient current is high, and energy is dissipated as heat ($I^2R$ loss) in the wires. Some energy can also be radiated away as electromagnetic waves.

Q: Does the amount of energy lost depend on the resistance of the connecting wires?

Surprisingly, no! The total amount of energy lost is completely independent of the wire's resistance. A higher resistance will just make the charge transfer slower, but the total heat dissipated will remain exactly the same.

Q: Why did we use $V_2 = 0$ in the formula?

The problem states that the second capacitor ($2.5\mu\text{F}$) is initially 'uncharged'. An uncharged capacitor has zero charge and therefore zero potential difference across its plates.

Q: What if the second capacitor was also charged, but with opposite polarity?

If the positive plate of one is connected to the negative plate of the other, you must assign a negative sign to one of the potentials. For example, if $V_2$ was $100\text{V}$ with opposite polarity, you would use $V_2 = -100\text{V}$ in the formula, making the term $(V_1 - (-V_2))^2 = (V_1 + V_2)^2$.

Q: The question says they are connected in 'series', but the formula looks like a parallel connection. Why?

When two capacitors are connected to each other in a closed loop, the connection is topologically identical to a parallel connection (they share the same two nodes). The term 'series' here just implies they are connected end-to-end in a single loop, but physically, charge redistributes until their voltages are equal, which is the defining characteristic of a parallel connection.

Question

A $5\mu	ext{F}$ capacitor is charged fully by a $220	ext{V}$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\mu	ext{F}$ capacitor. If the energy change during the charge redistribution is $\frac{X}{100}	ext{ J}$, then value of $X$ to the nearest integer is .........

Sigma Percentile · Accepted Answer

Imagine you have a bucket full of water and you suddenly connect it to an empty bucket using a pipe. Water will rush from the full bucket to the empty one until the water levels in both are exactly the same. During this chaotic rush, some energy is lost due to the friction of the water against the pipe.

This is exactly what happens in electrostatics when a charged capacitor is connected to an uncharged one. The 'water' is the electric charge, the 'water level' is the electric potential, and the 'friction' is the resistance of the connecting wires.

### The Direct Formula for Energy Loss

When two capacitors $C_1$ and $C_2$, initially at potentials $V_1$ and $V_2$, are connected together, charge redistributes until they reach a common potential. In this process, the total energy of the system decreases. The energy lost is dissipated as heat in the connecting wires.

While you could calculate the initial energy, find the common potential, calculate the final energy, and then subtract them, there is a much more elegant and direct formula that every JEE aspirant must know:

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

Notice how this formula looks remarkably similar to the formula for the loss of kinetic energy during a perfectly inelastic collision in mechanics: $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$. Physics is beautifully interconnected!

### Executing the Calculation

Let's plug our given values into this master equation. We have:
- $C_1 = 5\mu	ext{F}$
- $C_2 = 2.5\mu	ext{F}$
- $V_1 = 220	ext{V}$
- $V_2 = 0	ext{V}$ (since the second capacitor is uncharged)

Substituting these into our formula:

$$\Delta U = \frac{1}{2} \left( \frac{5 	imes 2.5}{5 + 2.5} ight) 	imes 10^{-6} 	imes (220 - 0)^2$$

Let's simplify the capacitance fraction first:

$$\frac{5 	imes 2.5}{7.5} = \frac{12.5}{7.5} = \frac{5}{3} \mu	ext{F}$$

Now, bringing it all together:

$$\Delta U = \frac{1}{2} 	imes \frac{5}{3} 	imes 10^{-6} 	imes (220)^2$$

$$\Delta U = \frac{5}{6} 	imes 10^{-6} 	imes 48400$$

$$\Delta U = \frac{242000}{6} 	imes 10^{-6} \approx 40333.33 	imes 10^{-6} 	ext{ J}$$

### Formatting the Final Answer

The question asks us to express the energy loss in the format $\frac{X}{100} 	ext{ J}$. Let's manipulate our result to match this structure:

$$\Delta U = 4.033 	imes 10^{-2} 	ext{ J} = \frac{4.033}{100} 	ext{ J}$$

Comparing this with $\frac{X}{100}$, we get $X = 4.033$.

Rounding off to the nearest integer, we arrive at our final answer:

**$X = 4$**

Always remember this direct formula. It not only saves precious minutes during the exam but also minimizes the chances of calculation errors that often occur in the longer, multi-step method.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Energy Loss Formula$

$Substituting Values$

$Simplifying Capacitance Term$

$Calculating Energy Loss$

$Formatting the Answer$

$The Way Forward$

The Sigma Insight: Capacitance and Capacitors

The Direct Formula for Energy Loss

Executing the Calculation

Formatting the Final Answer

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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