Irodov 3.133: The Heat Ignores the Working Cell – Amit & Deepali Sinha Foundation

Problem Irodov 3.133

A capacitor $C$ has one plate joined to the left side wire of a rectangular circuit and the other to the arm of a switch $Sw$ . The bottom wire holds, from left to right, a cell of emf $\mathcal{E}_1$ , a middle junction and a cell of emf $\mathcal{E}_2$ ; both cells have their positive terminal at the middle junction. In position $1$ the switch connects the capacitor to the right side wire, which runs down to the right end of the bottom wire; in position $2$ it connects the capacitor to the middle junction. How much heat is generated after the switch is moved from position $1$ to position $2$ ?

The circuit. The solid arrow shows the switch at contact 1, the dashed arrow the contact 2 it is moved to; both cells have the long (+) plate at the middle junction. — The circuit. The solid arrow shows the switch at contact $1$ , the dashed arrow the contact $2$ it is moved to; both cells have the long ( $+$ ) plate at the middle junction.

Given

$C$ , $\mathcal{E}_1$ , $\mathcal{E}_2$
steady state before the switch moves (assumed)

Find

the heat

Q

generated

Solution

Step 1The potentials of the circuit

Take the left side wire at $\varphi_L = 0$ . Crossing $\mathcal{E}_1$ from its negative plate on the left to its positive plate at the middle junction $M$ rises by $\mathcal{E}_1$ ; crossing $\mathcal{E}_2$ from its positive plate at $M$ to its negative plate at the right end $R$ falls by $\mathcal{E}_2$ :

\varphi_M = \mathcal{E}_1, \qquad \varphi_R = \mathcal{E}_1 - \mathcal{E}_2 \tag{1}

Step 2The charge on the capacitor in each position

Call $q$ the charge on the plate of $C$ attached to the switch arm. In position $1$ that plate is at $\varphi_R$ and in position $2$ at $\varphi_M$ , while the other plate stays at $0$ ; by (1):

q_1 = C(\mathcal{E}_1 - \mathcal{E}_2), \qquad q_2 = C\mathcal{E}_1 \tag{2}

Step 3Work done by the cells

The arm plate gains $q_2 - q_1$ , which flows from the left side wire through $\mathcal{E}_1$ , from its negative to its positive terminal, and up to the arm. With (2), the charge and the work of $\mathcal{E}_1$ on it are

\begin{align*} q_2 - q_1 &= C\mathcal{E}_1 - C(\mathcal{E}_1 - \mathcal{E}_2) = C\mathcal{E}_2 \\ A &= \mathcal{E}_1(q_2 - q_1) = C\mathcal{E}_1\mathcal{E}_2 \tag{3}\end{align*}

Step 4The change of stored energy

The energy of the capacitor before and after, from (2):

\begin{aligned} \Delta W &= \frac{q_2^2}{2C} - \frac{q_1^2}{2C} = \frac{1}{2}\,C\mathcal{E}_1^2 - \frac{1}{2}\,C(\mathcal{E}_1 - \mathcal{E}_2)^2\\[2pt] &= C\mathcal{E}_1\mathcal{E}_2 - \frac{1}{2}\,C\mathcal{E}_2^2 \end{aligned}

Step 5The heat

The work of the cell, (3), goes partly into the field and the rest into heat:

Q = A - \Delta W = C\mathcal{E}_1\mathcal{E}_2 - \left(C\mathcal{E}_1\mathcal{E}_2 - \frac{1}{2}\,C\mathcal{E}_2^2\right)

Q = \frac{1}{2}\,C\mathcal{E}_2^2

(4)

Check

With $\mathcal{E}_2 = 0$ both contacts sit at the same potential, the capacitor’s charge does not change, and no heat appears, as the answer says. With $\mathcal{E}_1 = 0$ the switch moves the capacitor from a voltage of magnitude $\mathcal{E}_2$ onto a plain wire: it discharges completely, and the heat must be its whole stored energy $C\mathcal{E}_2^2/2$ , which is the answer.

Remark

The answer is $(q_2 - q_1)^2/2C$ : the energy of the capacitor charged with the transferred charge alone. That is general. When a capacitor is switched from one fixed source to another, the heat is $C(\Delta U)^2/2$ , with $\Delta U$ the jump of voltage applied to it, whatever the source that does the work. Here the jump is $\mathcal{E}_2$ , which is why $\mathcal{E}_1$ , the very cell that drives the charge, drops out.