Irodov 3.199: Two Resistors Share the Stored Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.199

In the circuit drawn, a capacitor of capacitance $C = 5.00$ $\mu$ F is charged from a source of constant emf $\mathcal{E} = 200$ V through a resistance $R_2 = 330$ $\Omega$ while the two-way switch $Sw$ rests on contact $1$ . The switch is then thrown over from contact $1$ to contact $2$ , which joins the capacitor, $R_2$ and a resistance $R_1 = 500$ $\Omega$ in a closed loop without the source. How much heat is generated in $R_1$ ?

The circuit with the switch Sw on contact 1; the thin arrow shows it is thrown to contact 2. — The circuit with the switch $Sw$ on contact $1$ ; the thin arrow shows it is thrown to contact $2$ .

Given

$C = 5.00$ $\mu$ F
$\mathcal{E} = 200$ V
$R_1 = 500$ $\Omega$
$R_2 = 330$ $\Omega$
fully charged before the switch is thrown (assumed)

Find

the heat

Q_1

generated in

R_1

Solution

Step 1The charge before the switch is thrown

With the switch on contact $1$ , the loop is the source, $R_2$ and the capacitor; $R_1$ leads only to the idle contact $2$ and carries nothing. We take the switch to rest there long enough for the charging current to die away. Then no current flows through $R_2$ , and the capacitor holds the full emf, with the energy

W = \frac{C\mathcal{E}^2}{2} \tag{1}

Step 2The discharge after the switch is thrown

The switch on contact 2: the capacitor discharges round the loop through R_2 and R_1; the source hangs from the idle contact 1. — The switch on contact $2$ : the capacitor discharges round the loop through $R_2$ and $R_1$ ; the source hangs from the idle contact $1$ .

On contact $2$ the source is cut off, and the capacitor discharges round the loop through $R_2$ and $R_1$ in series. No source acts in this loop, so the whole stored energy (1) ends up as heat in the two resistances:

Q_1 + Q_2 = \frac{C\mathcal{E}^2}{2} \tag{2}

Step 3How the heat divides

At every instant the same current $I(t)$ flows through both resistances, so the Joule heats, integrated over the whole discharge, stand in the ratio of the resistances:

\begin{aligned} Q_1 &= R_1\int_0^\infty I^2\,dt\\[2pt] Q_2 &= R_2\int_0^\infty I^2\,dt\\[2pt] \frac{Q_1}{Q_1 + Q_2} &= \frac{R_1}{R_1 + R_2} \end{aligned}

Combining with (2), and numerically,

\begin{aligned} \frac{C\mathcal{E}^2}{2} &= \frac{5.00\times10^{-6}\cdot 200^2}{2}\ \text{J} = 0.100\ \text{J}\\[2pt] Q_1 &= 0.100\cdot\frac{500}{830}\ \text{J} \end{aligned}

Q_1 = \frac{C\mathcal{E}^2}{2}\,\frac{R_1}{R_1 + R_2} \approx 60\text{ mJ}

(3)

Check

Dimensions: $C\mathcal{E}^2$ is an energy and the resistance ratio is a pure number. Limits: with $R_2 = 0$ all the stored energy is released in $R_1$ , $Q_1 = C\mathcal{E}^2/2$ ; with $R_2 \gg R_1$ almost all of it goes to $R_2$ and $Q_1 \to 0$ . With $Q_1 = 0.0602$ J the heat in $R_2$ is $0.0398$ J, and $0.0398/0.0602 = 0.66$ , the ratio $330/500$ , as it should be.

Remark

The time course of the discharge never enters: whatever the shape of $I(t)$ , two resistances in series share the heat in the ratio of their resistances, so only the stored energy and that ratio matter. $R_2$ is also in the charging loop, where it takes another $C\mathcal{E}^2/2$ as heat while the source does work $C\mathcal{E}^2$ ; that heat is not part of the answer, since only $R_1$ is asked about and it carries no current then.