Irodov 3.177: The Capacitor Reads One Resistor – Amit & Deepali Sinha Foundation

Problem Irodov 3.177

Three branches join a wire on the left to a wire on the right. The upper branch holds, from left to right, a source of emf $\mathcal{E}_1 = 1.0$ V and a resistor $R_1 = 10~\Omega$ . The middle branch holds a second source of the same emf $\mathcal{E}_1$ and then a capacitor $C$ , whose left plate is $A$ and right plate $B$ . The lower branch holds a source of emf $\mathcal{E}_2 = 2.5$ V and a resistor $R_2 = 20~\Omega$ . Every source has its positive terminal facing right, and the internal resistances of the sources are negligible. Find the potential difference $\varphi_A - \varphi_B$ between the capacitor plates.

The circuit. All three cells have their long (+) plate on the right; A is the capacitor plate on the cell side. — The circuit. All three cells have their long ( $+$ ) plate on the right; $A$ is the capacitor plate on the cell side.

Given

$\mathcal{E}_1 = 1.0$ V (upper and middle sources)
$\mathcal{E}_2 = 2.5$ V
$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
internal resistances negligible
steady state (assumed)

Find

\varphi_A - \varphi_B

Solution

Step 1Only the outer loop carries current

We take the steady state, once the capacitor has charged. No current then passes the capacitor, so the middle branch carries none, and the upper and lower branches form a single loop with one current $I$ . Take $I$ clockwise, as drawn: rightward through $R_1$ and leftward through $R_2$ .

The current I of the outer loop, taken clockwise. The middle branch carries no current. — The current $I$ of the outer loop, taken clockwise. The middle branch carries no current.

Going clockwise, $\mathcal{E}_1$ is crossed from $-$ to $+$ and $\mathcal{E}_2$ from $+$ to $-$ , so the loop rule reads

\mathcal{E}_1 - \mathcal{E}_2 = I(R_1 + R_2)

I = \frac{\mathcal{E}_1 - \mathcal{E}_2}{R_1 + R_2}

(1)

Step 2The two plates against the two wires

Call $\varphi_L$ and $\varphi_R$ the potentials of the left and right wires. Plate $A$ is reached from the left wire across the middle cell, from $-$ to $+$ , with no current anywhere in that branch; plate $B$ is joined directly to the right wire.

\varphi_A = \varphi_L + \mathcal{E}_1, \qquad \varphi_B = \varphi_R

Along the upper branch, from the left wire to the right one, the potential rises by $\mathcal{E}_1$ across the cell and falls by $IR_1$ through the resistor:

\varphi_R = \varphi_L + \mathcal{E}_1 - IR_1

Subtracting this from the expression for $\varphi_A$ , the $\varphi_L + \mathcal{E}_1$ cancels:

\varphi_A - \varphi_B = IR_1

Substituting (1) and then the numbers,

\varphi_A - \varphi_B = \frac{(1.0 - 2.5)(10)}{10 + 20}~\text{V}

which gives

\varphi_A - \varphi_B = \frac{(\mathcal{E}_1 - \mathcal{E}_2)R_1}{R_1 + R_2} = -0.50\text{ V}

(2)

Plate $B$ is $0.50$ V above plate $A$ : $B$ carries the positive charge. The current $I = -0.050$ A is negative, so it actually circulates anticlockwise, driven by the larger $\mathcal{E}_2$ .

Check

The right wire can be reached along the lower branch instead. From the left wire, across $\mathcal{E}_2$ from $-$ to $+$ the potential rises by $2.5$ V, and through $R_2$ , walked from left to right against the drawn $I$ , it rises by $IR_2 = (-0.050)(20) = -1.0$ V; so $\varphi_R = \varphi_L + 1.5$ V. Then $\varphi_A - \varphi_B = \mathcal{E}_1 - 1.5~\text{V} = -0.50$ V, the same. Special case: with $\mathcal{E}_1 = \mathcal{E}_2$ no current flows and the capacitor stays uncharged, as the formula gives.

Remark

The capacitance never enters. Because the middle cell repeats the upper one, plate $A$ sits at the same potential as the junction between $\mathcal{E}_1$ and $R_1$ in the upper branch, so the capacitor simply reads the voltage across $R_1$ . The capacitor branch is a voltmeter that draws no current, which is why it disturbs nothing in the loop.