Irodov 3.184: A Capacitor Reads Two Cells – Amit & Deepali Sinha Foundation

Problem Irodov 3.184

A circuit is a rectangle whose left and right sides are plain wires, with a horizontal rung across its middle. The top wire holds, from left to right, a resistor $R_3$ and a capacitor $C$ whose left plate is $A$ and right plate is $B$ . The rung holds, from left to right, a resistor $R_2$ , a middle junction and a resistor $R_1$ . A source of emf $\mathcal{E}_2$ joins the rung’s middle junction to the junction between $R_3$ and the capacitor, with its positive terminal toward the capacitor. The bottom wire holds a source of emf $\mathcal{E}_1$ with its positive terminal on the right. Given $\mathcal{E}_1 = 4.0$ V, $\mathcal{E}_2 = 1.0$ V, $R_1 = 10$ $\Omega$ , $R_2 = 20$ $\Omega$ and $R_3 = 30$ $\Omega$ , and neglecting the internal resistances of the sources, find the potential difference $\varphi_A - \varphi_B$ between the capacitor plates.

The circuit. Both cells are drawn with the long (+) plate: E_2 up toward plate A, E_1 to the right. Plate A is on the left of C, plate B on the right. — The circuit. Both cells are drawn with the long ( $+$ ) plate: $\mathcal{E}_2$ up toward plate $A$ , $\mathcal{E}_1$ to the right. Plate $A$ is on the left of $C$ , plate $B$ on the right.

Given

$\mathcal{E}_1 = 4.0$ V
$\mathcal{E}_2 = 1.0$ V
$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
$R_3 = 30~\Omega$
internal resistances negligible
steady state (assumed)

Find

\varphi_A - \varphi_B

Solution

Step 1What the capacitor does in the steady state

Once the capacitor has charged, which we assume has happened, no current passes through it, so the top wire carries current only through $R_3$ , and that current goes on down through $\mathcal{E}_2$ into the rung. Call $J$ the junction between $R_3$ and $C$ , and $M$ the rung’s middle junction. Take the left side of the rectangle as the zero of potential. The source $\mathcal{E}_1$ , with no internal resistance, then holds the right side at $\mathcal{E}_1$ :

\varphi_{\mathrm{left}} = 0, \qquad \varphi_{\mathrm{right}} = \mathcal{E}_1 \tag{1}

Plate $A$ sits on $J$ and plate $B$ on the right side. Crossing $\mathcal{E}_2$ from $M$ to its positive terminal at $J$ raises the potential by $\mathcal{E}_2$ :

\varphi_A = \varphi_J = \varphi_M + \mathcal{E}_2, \qquad \varphi_B = \mathcal{E}_1 \tag{2}

Step 2Kirchhoff’s current rule at the junction

M

Three currents flow into $M$ : through $R_2$ from the left side, through $R_1$ from the right side, and down through $\mathcal{E}_2$ , which carries the current of $R_3$ flowing from the left side toward $J$ . Their sum vanishes. With (1) and (2):

\frac{0 - \varphi_M}{R_2} + \frac{\mathcal{E}_1 - \varphi_M}{R_1} + \frac{0 - (\varphi_M + \mathcal{E}_2)}{R_3} = 0

Collecting the terms in $\varphi_M$ and multiplying through by $R_1R_2R_3$ :

\begin{aligned} \varphi_M\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right) &= \frac{\mathcal{E}_1}{R_1} - \frac{\mathcal{E}_2}{R_3}\\[2pt] \varphi_M\,(R_1R_2 + R_2R_3 + R_3R_1) &= \mathcal{E}_1R_2R_3 - \mathcal{E}_2R_1R_2 \end{aligned}

\varphi_M = \frac{\mathcal{E}_1R_2R_3 - \mathcal{E}_2R_1R_2}{R_1R_2 + R_2R_3 + R_3R_1}

(3)

Step 3The voltage across the plates

By (2), the difference between the plates is $\varphi_M + \mathcal{E}_2 - \mathcal{E}_1$ . Writing it over the denominator of (3), with $S = R_1R_2 + R_2R_3 + R_3R_1$ :

\begin{aligned} \varphi_A - \varphi_B &= \frac{\mathcal{E}_1R_2R_3 - \mathcal{E}_2R_1R_2 + (\mathcal{E}_2 - \mathcal{E}_1)(R_1R_2 + R_2R_3 + R_3R_1)}{S}\\[2pt] &= \frac{\mathcal{E}_2(R_2R_3 + R_3R_1) - \mathcal{E}_1(R_1R_2 + R_3R_1)}{S} \end{aligned}

In the numerator $R_3$ factors out of the first bracket and $R_1$ out of the second:

\mathcal{E}_2(R_2R_3 + R_3R_1) - \mathcal{E}_1(R_1R_2 + R_3R_1) = \mathcal{E}_2R_3(R_1 + R_2) - \mathcal{E}_1R_1(R_2 + R_3)

Numerically, the denominator and the numerator are

\begin{aligned} S &= 200 + 600 + 300 = 1100~\Omega^2\\[2pt] \mathcal{E}_2R_3(R_1 + R_2) - \mathcal{E}_1R_1(R_2 + R_3) &= 1.0 \cdot 30 \cdot 30 - 4.0 \cdot 10 \cdot 50 = -1100\text{ V}\,\Omega^2 \end{aligned}

so that

\varphi_A - \varphi_B = \frac{\mathcal{E}_2R_3(R_1 + R_2) - \mathcal{E}_1R_1(R_2 + R_3)}{R_1R_2 + R_2R_3 + R_3R_1} = -1.0\text{ V}

(4)

Plate $B$ is the positive one.

Check

Two limits of (4). As $R_3 \to 0$ , the junction $J$ is wired straight to the left side, so $\varphi_A = 0$ and the answer must be $-\mathcal{E}_1$ ; the formula gives $-\mathcal{E}_1R_1R_2/(R_1R_2) = -\mathcal{E}_1$ . As $R_3 \to \infty$ , no current enters the $\mathcal{E}_2$ branch, the rung is a plain divider with $\varphi_M = \mathcal{E}_1R_2/(R_1 + R_2)$ , and $\varphi_A - \varphi_B = \mathcal{E}_2 - \mathcal{E}_1R_1/(R_1 + R_2)$ ; dividing the numerator and the denominator of the formula by $R_3$ and letting $R_3 \to \infty$ gives $[\mathcal{E}_2(R_1 + R_2) - \mathcal{E}_1R_1]/(R_1 + R_2)$ , the same. With the numbers, $\varphi_M = (4 \cdot 600 - 1 \cdot 200)/1100 = 2.0$ V, so $\varphi_A = 3.0$ V against $\varphi_B = 4.0$ V.

Remark

The capacitor drops out of the steady state completely: it only reads the potential difference between two points of a resistive network, as an ideal voltmeter would, and its capacitance enters only the charge, $q = C\,|\varphi_A - \varphi_B|$ . The node equation (3) is Millman’s form, the potential of a node as the conductance-weighted mean of what each branch would drive it to: $\mathcal{E}_1$ through $R_1$ , $0$ through $R_2$ , and $-\mathcal{E}_2$ through $R_3$ .