Irodov 3.186: A Bare Wire Across the Bridge – Amit & Deepali Sinha Foundation

Problem Irodov 3.186

A bridge network joins the terminal $A$ to the terminal $B$ along two paths: the upper one through $R_1$ to the point $C$ and then through $R_2$ , the lower one through $R_3$ to the point $D$ and then through $R_4$ . The points $C$ and $D$ are joined by a plain wire of negligible resistance. A constant voltage $V = 25$ V is maintained between $A$ and $B$ , with $A$ the positive terminal. Given $R_1 = 1.0$ $\Omega$ , $R_2 = 2.0$ $\Omega$ , $R_3 = 3.0$ $\Omega$ and $R_4 = 4.0$ $\Omega$ , find the magnitude and the direction of the current in the wire $CD$ .

The bridge. A is the positive terminal and B the negative one; R_1 and R_2 form the path through C, R_3 and R_4 the path through D, and CD is a bare wire. — The bridge. $A$ is the positive terminal and $B$ the negative one; $R_1$ and $R_2$ form the path through $C$ , $R_3$ and $R_4$ the path through $D$ , and $CD$ is a bare wire.

Given

$V = 25$ V, $A$ positive
$R_1 = 1.0~\Omega$
$R_2 = 2.0~\Omega$
$R_3 = 3.0~\Omega$
$R_4 = 4.0~\Omega$
$CD$ of zero resistance

Find

the current in

CD

, magnitude and direction

Solution

Step 1The bare wire makes two parallel pairs

With $C$ and $D$ joined, $R_1$ and $R_3$ both run from $A$ to that node, and $R_2$ and $R_4$ both run from it to $B$ .

The same network redrawn: R_1 and R_3 both run from A to the wire CD, and R_2 and R_4 from it to B, two parallel pairs in series. — The same network redrawn: $R_1$ and $R_3$ both run from $A$ to the wire $CD$ , and $R_2$ and $R_4$ from it to $B$ , two parallel pairs in series.

The two parallel pairs have the resistances

R_{13} = \frac{R_1R_3}{R_1 + R_3}, \qquad R_{24} = \frac{R_2R_4}{R_2 + R_4} \tag{1}

and the voltage $V$ divides between them in proportion to these resistances, so the drop from $A$ to the node is

U = V\,\frac{R_{13}}{R_{13} + R_{24}} \tag{2}

Step 2The currents in the four resistors

The drop $U$ lies across $R_1$ and $R_3$ , the rest $V - U$ across $R_2$ and $R_4$ . Substituting (1) into (2) and writing $D_0 = R_1R_3(R_2 + R_4) + R_2R_4(R_1 + R_3)$ :

U = \frac{V R_1R_3(R_2 + R_4)}{D_0}, \qquad V - U = \frac{V R_2R_4(R_1 + R_3)}{D_0}

The current in $R_1$ (from $A$ toward $C$ ) and the current in $R_2$ (from $C$ toward $B$ ) follow by Ohm’s law:

I_1 = \frac{U}{R_1} = \frac{V R_3(R_2 + R_4)}{D_0}, \qquad I_2 = \frac{V - U}{R_2} = \frac{V R_4(R_1 + R_3)}{D_0}

Step 3The current in

CD

from the rule at

C

At $C$ the current $I_1$ arrives and $I_2$ leaves through $R_2$ ; the difference must leave through the wire toward $D$ . Counting the current in $CD$ positive from $C$ to $D$ :

I_{CD} = I_1 - I_2 = \frac{V\,[R_2R_3 + R_3R_4 - R_1R_4 - R_3R_4]}{D_0}

The terms $R_3R_4$ cancel. Numerically,

D_0 = 3 \cdot 6 + 8 \cdot 4 = 50~\Omega^3, \qquad R_2R_3 - R_1R_4 = 6 - 4 = 2~\Omega^2

so that

I_{CD} = \frac{V(R_2R_3 - R_1R_4)}{R_1R_3(R_2 + R_4) + R_2R_4(R_1 + R_3)} = 1.0\text{ A}

(3)

The value is positive, so the current of $1.0$ A flows in the wire from $C$ to $D$ .

Check

The numbers in full: $R_{13} = 0.75~\Omega$ and $R_{24} = 1.33~\Omega$ , so the total current from $A$ is $12$ A, with $U = 9.0$ V and $V - U = 16$ V. Then $I_1 = 9.0$ A and $I_3 = 3.0$ A arrive at the joined node, while $I_2 = 8.0$ A and $I_4 = 4.0$ A leave it: $C$ receives $9$ A and passes on $8$ A, so $1$ A crosses to $D$ , which receives $3$ A and must pass on $4$ A. Both junctions balance. A special case: when $R_2R_3 = R_1R_4$ , (3) gives zero, which is the balance condition of the bridge, where $C$ and $D$ would be at one potential even without the wire.

Remark

The sign of $R_2R_3 - R_1R_4$ alone decides the direction. Without the wire, $C$ would sit at the fraction $R_2/(R_1 + R_2)$ of $V$ above $B$ and $D$ at $R_4/(R_3 + R_4)$ ; here those are $2/3$ and $4/7$ , so $C$ would be the higher point, and the wire relieves that difference by carrying current from $C$ to $D$ . A wire across a bridge is the extreme case of a galvanometer of zero resistance.