Irodov 3.181: Three Branches, One Node Equation – Amit & Deepali Sinha Foundation

Problem Irodov 3.181

Three branches join a wire on the left to a wire on the right. The upper branch holds, from left to right, a resistor $R_2 = 20~\Omega$ and a source of emf $\mathcal{E}_2 = 3.7$ V whose positive terminal faces left, toward $R_2$ . The middle branch holds a source of emf $\mathcal{E}_1 = 1.5$ V whose positive terminal faces right, and then a resistor $R_1 = 10~\Omega$ . The lower branch holds a resistor $R = 5.0~\Omega$ alone. The internal resistances of the sources are negligible. Find the magnitude and direction of the current through $R$ .

The circuit. E_2 has its long (+) plate on the left, facing R_2; E_1 has its long (+) plate on the right, facing R_1. — The circuit. $\mathcal{E}_2$ has its long ( $+$ ) plate on the left, facing $R_2$ ; $\mathcal{E}_1$ has its long ( $+$ ) plate on the right, facing $R_1$ .

Given

$\mathcal{E}_1 = 1.5$ V
$\mathcal{E}_2 = 3.7$ V
$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
$R = 5.0~\Omega$
internal resistances negligible

Find

the current through

R

, magnitude and direction

Solution

Step 1Each branch current from the voltage across the wires

The circuit has only two junctions, the left and right wires, so one unknown fixes everything: the voltage $U = \varphi_L - \varphi_R$ between them. Take every branch current from left to right, as drawn.

The three branch currents, each taken from left to right.

Walking each branch from left to right with its current: in the upper branch the potential falls by $I_2R_2$ and then by $\mathcal{E}_2$ (crossed from $+$ to $-$ ); in the middle it rises by $\mathcal{E}_1$ and falls by $I_1R_1$ ; in the lower it falls by $IR$ .

\begin{aligned} U &= I_2R_2 + \mathcal{E}_2\\ U &= I_1R_1 - \mathcal{E}_1\\ U &= IR \end{aligned}

Solved for the currents:

\begin{aligned} I_2 &= \frac{U - \mathcal{E}_2}{R_2}\\[2pt] I_1 &= \frac{U + \mathcal{E}_1}{R_1}\\[2pt] I &= \frac{U}{R} \end{aligned} \tag{1}

Step 2The junction rule fixes the voltage

All three currents arrive at the right-hand wire, and nothing else leaves it, so they add up to zero. Using (1):

\frac{U - \mathcal{E}_2}{R_2} + \frac{U + \mathcal{E}_1}{R_1} + \frac{U}{R} = 0

Multiplying through by $RR_1R_2$ and collecting the terms in $U$ :

U(R_1R_2 + RR_1 + RR_2) = R(R_1\mathcal{E}_2 - R_2\mathcal{E}_1)

Step 3The current through

R

The current through $R$ is $U/R$ , by (1), so dividing both sides by $R$ and by the bracket gives it directly. Numerically, first,

I = \frac{(10)(3.7) - (20)(1.5)}{(5.0)(10) + (10)(20) + (20)(5.0)}~\text{A} = \frac{7.0}{350}~\text{A}

and so

I = \frac{R_1\mathcal{E}_2 - R_2\mathcal{E}_1}{RR_1 + R_1R_2 + R_2R} = 0.020\text{ A}

(2)

The value is positive, so the current of $20$ mA flows through $R$ from left to right, as drawn.

Check

With $U = IR = 0.10$ V the other two branches give $I_1 = (0.10 + 1.5)/10 = 0.16$ A and $I_2 = (0.10 - 3.7)/20 = -0.18$ A, and the three currents into the right-hand wire add to $0.16 - 0.18 + 0.02 = 0$ , as the junction rule demands. Special case: when $\mathcal{E}_2/R_2 = \mathcal{E}_1/R_1$ the two sources balance and no current passes $R$ , as the numerator shows.

Remark

With the two wires shorted, each source would drive its own current through its series resistor: $\mathcal{E}_1/R_1 = 0.15$ A to the right through the middle branch and $\mathcal{E}_2/R_2 = 0.185$ A to the left through the top. The current in $R$ comes from the small difference of these two nearly equal pushes, which is why it is so much smaller than either. Choosing one node and writing every branch through its voltage, as here, needs one equation where loop currents would need two.