Irodov 3.174: Walking the Loop Between Two Points – Amit & Deepali Sinha Foundation

Problem Irodov 3.174

A single closed loop carries, along its upper side from left to right, a resistor $R_1 = 10~\Omega$ and a source of emf $\mathcal{E}_1 = 5.0$ V whose positive terminal faces right. Along its lower side, again from left to right, it carries a source of emf $\mathcal{E}_2 = 2.0$ V whose positive terminal also faces right, and a resistor $R_2 = 20~\Omega$ . Point $1$ lies on the left side of the loop and point $2$ on the right side. The internal resistances of the sources are negligible. Find the potential difference $\varphi_1 - \varphi_2$ .

The loop. Both cells have their long (+) plate on the right, so going round the loop they act against each other. — The loop. Both cells have their long ( $+$ ) plate on the right, so going round the loop they act against each other.

Given

$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
$\mathcal{E}_1 = 5.0$ V
$\mathcal{E}_2 = 2.0$ V
internal resistances negligible

Find

\varphi_1 - \varphi_2

Solution

Step 1One current round the loop

The circuit has no junction, so one current $I$ flows everywhere in it. Take $I$ clockwise, as drawn below. Going clockwise, $\mathcal{E}_1$ is crossed from its $-$ plate to its $+$ plate and helps the current, while $\mathcal{E}_2$ is crossed from $+$ to $-$ and opposes it.

The current I taken clockwise, with the gray sense of the loop. — The current $I$ taken clockwise, with the gray sense of the loop.

Kirchhoff’s loop rule, the sum of the emfs equal to the sum of the $IR$ drops, reads

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

so that

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

(1)

Numerically,

I = \frac{5.0 - 2.0}{10 + 20}~\text{A} = 0.10~\text{A}

It is positive, so the current really is clockwise.

Step 2Walking from 1 to 2 along the top

Go from point $1$ up the left side and along the top to point $2$ , in the sense of the current. Through $R_1$ the potential falls by $IR_1$ , since current flows from high potential to low; across $\mathcal{E}_1$ , from $-$ to $+$ , it rises by $\mathcal{E}_1$ .

\varphi_2 = \varphi_1 - IR_1 + \mathcal{E}_1

Rearranging:

\varphi_1 - \varphi_2 = IR_1 - \mathcal{E}_1

Substituting (1) and the numbers,

\varphi_1 - \varphi_2 = (0.10~\text{A})(10~\Omega) - 5.0~\text{V}

\varphi_1 - \varphi_2 = \frac{(\mathcal{E}_1 - \mathcal{E}_2)R_1}{R_1 + R_2} - \mathcal{E}_1 = -4.0\text{ V}

(2)

Point $2$ is $4.0$ V above point $1$ .

Check

The potential difference cannot depend on the path. Going instead from $1$ down and along the bottom to $2$ : across $\mathcal{E}_2$ , from $-$ to $+$ , the potential rises by $\mathcal{E}_2 = 2.0$ V, and through $R_2$ , walked against the current, it rises by $IR_2 = 2.0$ V. So $\varphi_2 = \varphi_1 + 4.0$ V, the same result.

Remark

Points $1$ and $2$ are joined by two branches, each a source in series with a resistor. Over the common denominator the answer is $\varphi_2 - \varphi_1 = (\mathcal{E}_1R_2 + \mathcal{E}_2R_1)/(R_1 + R_2)$ , the mean of the two emfs weighted by the conductances $1/R_1$ and $1/R_2$ : here $(5.0 \cdot 20 + 2.0 \cdot 10)/30 = 4.0$ V, between $2.0$ and $5.0$ V and nearer to the source with the smaller resistance in series.