Thermodynamics of Circuits: Charge Redistribution & Heat Loss

When you connect two charged capacitors together with a conducting wire, charge will immediately flow from the one at a higher potential to the one at a lower potential. This flow of charge is a transient electric current.

Eventually, the flow stops. This happens when both capacitors reach the exact same electrical pressure, known as the Common Potential ($V_{common}$).

However, this process is never perfectly efficient. The sudden rush of charge creates a spark and heats up the connecting wires. Even if we assume the wires have zero resistance, the energy must be lost as an electromagnetic pulse.

Step 1: Conservation of Charge and Common Potential

Consider two isolated capacitors:

• Capacitor 1: Capacitance $C_1$, initially charged to potential $V_1$. Its initial charge is $Q_1 = C_1 V_1$.
• Capacitor 2: Capacitance $C_2$, initially charged to potential $V_2$. Its initial charge is $Q_2 = C_2 V_2$.

We connect their positive terminals together and their negative terminals together (parallel connection).

According to the Law of Conservation of Charge, the total charge before connection must equal the total charge after connection:

$Q_{total} = Q_1 + Q_2 = C_1 V_1 + C_2 V_2$

After the redistribution, they share the same common potential $V_{common}$. They now act as a single equivalent capacitor $C_{eq} = C_1 + C_2$.

We can find the common potential using the fundamental definition $V = Q/C$:

$V_{common} = \frac{Q_{total}}{C_{eq}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

Step 2: The Energy Audit

To find out how much energy was burned off as heat, we must calculate the total energy of the system before the connection ($U_i$) and after the connection ($U_f$).

Initial Energy: $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$

Final Energy: $U_f = \frac{1}{2} (C_1 + C_2) V_{common}^2$

Substitute our expression for $V_{common}$ into the final energy equation:

$U_f = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \right)^2$ $U_f = \frac{1}{2} \frac{(C_1 V_1 + C_2 V_2)^2}{C_1 + C_2}$

Step 3: Calculating the Heat Dissipated ($\Delta H$)

The heat produced is simply the missing energy: $\Delta H = U_i - U_f$.

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

Finding a common denominator and expanding the numerators algebraically:

$\Delta H = \frac{1}{2(C_1 + C_2)} \left[ (C_1 + C_2)(C_1 V_1^2 + C_2 V_2^2) - (C_1^2 V_1^2 + C_2^2 V_2^2 + 2C_1 C_2 V_1 V_2) \right]$

When you expand the first bracket, the $C_1^2 V_1^2$ and $C_2^2 V_2^2$ terms perfectly cancel out, leaving only terms with $C_1 C_2$:

$\Delta H = \frac{1}{2(C_1 + C_2)} \left[ C_1 C_2 V_1^2 + C_1 C_2 V_2^2 - 2C_1 C_2 V_1 V_2 \right]$

Factor out the $C_1 C_2$:

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

This reveals a perfect square binomial, giving us the master formula for heat loss:

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

The Mechanics Parallel: Notice the striking similarity to the formula for kinetic energy lost in a perfectly inelastic collision: $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 - u_2)^2$. In mechanics, mass dictates inertia and velocity dictates state; in electrostatics, capacitance dictates inertia and voltage dictates state.

The JEE Trap: Connecting Opposite Polarities

If a problem states that the positive plate of $C_1$ is connected to the negative plate of $C_2$, the initial total charge is not a simple sum. They partially cancel out.

The common potential becomes: $V_{common} = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2}$

Consequently, the $(V_1 - V_2)^2$ term in the heat loss formula becomes $(V_1 + V_2)^2$. Connecting opposite polarities forces a much larger charge redistribution, resulting in a significantly more violent spark and greater heat loss.