Irodov 3.185: The Star Point Weighs Its Neighbours – Amit & Deepali Sinha Foundation

Problem Irodov 3.185

Three resistors meet at a common junction $0$ in a star: $R_1$ joins it to the terminal $1$ , $R_2$ to the terminal $2$ , and $R_3$ to the terminal $3$ . The terminals are held at the potentials $\varphi_1 = 10$ V, $\varphi_2 = 6$ V and $\varphi_3 = 5$ V, and $R_1 = 10$ $\Omega$ , $R_2 = 20$ $\Omega$ , $R_3 = 30$ $\Omega$ . Find the current through $R_1$ .

Given

$R_1 = 10~\Omega$
$R_2 = 20~\Omega$
$R_3 = 30~\Omega$
$\varphi_1 = 10$ V
$\varphi_2 = 6$ V
$\varphi_3 = 5$ V

Find

the current

I_1

through

R_1

Solution

Step 1The potential of the centre

The centre $0$ is a junction with nothing else attached, so the currents flowing into it along the three arms add up to zero. The current in arm $k$ , counted toward $0$ , is $(\varphi_k - \varphi_0)/R_k$ :

\frac{\varphi_1 - \varphi_0}{R_1} + \frac{\varphi_2 - \varphi_0}{R_2} + \frac{\varphi_3 - \varphi_0}{R_3} = 0

Solving for $\varphi_0$ :

\varphi_0 = \frac{\varphi_1/R_1 + \varphi_2/R_2 + \varphi_3/R_3}{1/R_1 + 1/R_2 + 1/R_3}

(1)

Numerically,

\varphi_0 = \frac{1.00 + 0.30 + 0.167}{0.100 + 0.050 + 0.0333}\text{ V} = \frac{1.467}{0.1833}\text{ V} = 8.0\text{ V}

Step 2The current through

R_1

Counting the current from terminal $1$ toward $0$ , Ohm’s law for the arm gives $I_1 = (\varphi_1 - \varphi_0)/R_1$ . Subtracting (1) from $\varphi_1$ over the common denominator, the $\varphi_1/R_1$ terms cancel:

\varphi_1 - \varphi_0 = \frac{(\varphi_1 - \varphi_2)/R_2 + (\varphi_1 - \varphi_3)/R_3}{1/R_1 + 1/R_2 + 1/R_3}

Dividing by $R_1$ :

I_1 = \frac{(\varphi_1 - \varphi_2)/R_2 + (\varphi_1 - \varphi_3)/R_3}{1 + R_1/R_2 + R_1/R_3}

Multiplying the numerator and the denominator by $R_2R_3$ clears the inner fractions. Numerically, the new numerator and denominator are

\begin{aligned} R_3(\varphi_1 - \varphi_2) + R_2(\varphi_1 - \varphi_3) &= 30 \cdot 4 + 20 \cdot 5 = 220\text{ V}\,\Omega\\[2pt] R_1R_2 + R_2R_3 + R_3R_1 &= 200 + 600 + 300 = 1100~\Omega^2 \end{aligned}

so that

I_1 = \frac{R_3(\varphi_1 - \varphi_2) + R_2(\varphi_1 - \varphi_3)}{R_1R_2 + R_2R_3 + R_3R_1} = 0.20\text{ A}

(2)

The current is positive, so it flows from terminal $1$ into the centre.

Check

With $\varphi_0 = 8.0$ V the three arm currents toward $0$ are $(10 - 8)/10 = 0.20$ A, $(6 - 8)/20 = -0.10$ A and $(5 - 8)/30 = -0.10$ A. They sum to zero, as the current rule at $0$ requires: the $0.20$ A arriving through $R_1$ leaves as $0.10$ A through each of the other arms. A limit: if all three terminals are at one potential, both differences in (2) vanish and no current flows, as it must.

Remark

The centre’s potential (1) is the mean of the terminal potentials weighted by the arm conductances $1/R_k$ , so it always lies between the lowest and the highest of them, and a low-resistance arm pulls it toward its own terminal. Here $\varphi_0 = 8.0$ V sits nearer the $10$ V terminal because $R_1$ is the smallest resistor. The same weighted mean gives the potential of any node joined to fixed potentials through resistors alone.