Irodov 3.182: Three Sources Between Two Nodes – Amit & Deepali Sinha Foundation

Problem Irodov 3.182

Two junctions, $A$ above and $B$ below, are joined by three branches. The first holds a source of emf $\mathcal{E}_1 = 1.5$ V, positive terminal toward $A$ , in series with a resistor $R_1 = 10~\Omega$ . The second holds a resistor $R_2 = 20~\Omega$ in series with a source of emf $\mathcal{E}_2 = 2.0$ V, positive terminal toward $A$ . The third holds a source of emf $\mathcal{E}_3 = 2.5$ V, positive terminal toward $B$ , in series with a resistor $R_3 = 30~\Omega$ . The internal resistances of the sources are negligible. Find:

the current through $R_1$
the potential difference $\varphi_A - \varphi_B$

The circuit. E_1 and E_2 have their long (+) plates up, toward A; E_3 has its long (+) plate down, toward B. — The circuit. $\mathcal{E}_1$ and $\mathcal{E}_2$ have their long ( $+$ ) plates up, toward $A$ ; $\mathcal{E}_3$ has its long ( $+$ ) plate down, toward $B$ .

Given

$\mathcal{E}_1 = 1.5$ V
$\mathcal{E}_2 = 2.0$ V
$\mathcal{E}_3 = 2.5$ V
$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
$R_3 = 30~\Omega$
internal resistances negligible

Find

(a)

I_1

through

R_1

· (b)

\varphi_A - \varphi_B

Solution

Step 1Every branch sets the same voltage

Take each branch current upward, from $B$ to $A$ , as drawn, and call $U = \varphi_A - \varphi_B$ . Walking up each branch with its current, the potential falls by the $IR$ drop and rises by an emf crossed from $-$ to $+$ , or falls by one crossed from $+$ to $-$ , as $\mathcal{E}_3$ is.

The three branch currents, each taken upward from B to A. — The three branch currents, each taken upward from $B$ to $A$ .

\begin{aligned} U &= \mathcal{E}_1 - I_1R_1\\ U &= \mathcal{E}_2 - I_2R_2\\ U &= -\mathcal{E}_3 - I_3R_3 \end{aligned} \tag{1}

All three currents arrive at $A$ , and nothing else leaves it:

I_1 + I_2 + I_3 = 0 \tag{2}

Step 2(a) The current through

R_1

Equating the first line of (1) with the second and with the third gives the other two currents in terms of $I_1$ :

\begin{aligned} I_2 &= \frac{\mathcal{E}_2 - \mathcal{E}_1 + I_1R_1}{R_2}\\[2pt] I_3 &= \frac{-\mathcal{E}_3 - \mathcal{E}_1 + I_1R_1}{R_3} \end{aligned}

Putting both into (2) and multiplying through by $R_2R_3$ :

I_1R_2R_3 + (\mathcal{E}_2 - \mathcal{E}_1 + I_1R_1)R_3 - (\mathcal{E}_1 + \mathcal{E}_3 - I_1R_1)R_2 = 0

Collecting the terms in $I_1$ :

I_1(R_1R_2 + R_2R_3 + R_3R_1) = R_3(\mathcal{E}_1 - \mathcal{E}_2) + R_2(\mathcal{E}_1 + \mathcal{E}_3)

Dividing by the bracket gives $I_1$ . Numerically, first,

I_1 = \frac{(30)(1.5 - 2.0) + (20)(1.5 + 2.5)}{(10)(20) + (20)(30) + (30)(10)}~\text{A} = \frac{65}{1100}~\text{A}

and so

I_1 = \frac{R_3(\mathcal{E}_1 - \mathcal{E}_2) + R_2(\mathcal{E}_1 + \mathcal{E}_3)}{R_1R_2 + R_2R_3 + R_3R_1} \approx 0.059\text{ A}

(3)

The value is positive: about $0.06$ A flows up through $R_1$ , from $B$ toward $A$ .

Step 3(b) The voltage between

A

and

B

The first line of (1), with $I_1$ from (3), gives $U$ . Numerically,

U = 1.5~\text{V} - (0.0591~\text{A})(10~\Omega)

which gives

\varphi_A - \varphi_B = \mathcal{E}_1 - I_1R_1 \approx 0.91\text{ V}

(4)

Check

The other two branches must agree. With $U = 0.909$ V, the second line of (1) gives $I_2 = (2.0 - 0.909)/20 = 0.0545$ A and the third $I_3 = (-2.5 - 0.909)/30 = -0.1136$ A, and $0.0591 + 0.0545 - 0.1136 = 0$ , as (2) requires. In exact form $U = 10/11$ V.

Remark

Eliminating the currents instead gives $U$ as a weighted mean, $U = (\mathcal{E}_1/R_1 + \mathcal{E}_2/R_2 - \mathcal{E}_3/R_3)/(1/R_1 + 1/R_2 + 1/R_3)$ : each branch pulls the common voltage toward its own emf, signed by which way it faces, with a weight equal to its conductance. The reversed $\mathcal{E}_3$ pulls toward $-2.5$ V, which is why $\mathcal{E}_1$ and $\mathcal{E}_3$ enter the current through $R_1$ as a sum.