Irodov 3.187: Turning the Crossed Bridge Around – Amit & Deepali Sinha Foundation

Problem Irodov 3.187

Two paths join the terminal $A$ to the terminal $B$ . The upper path is a resistor $r$ followed by a resistor $R$ ; the lower path is a resistor $R$ followed by a resistor $r$ . A further resistor $r$ joins the junction in the middle of the upper path to the junction in the middle of the lower path. Find the resistance of the network between $A$ and $B$ .

The network. The upper path is r then R, the lower path R then r, and a third r joins their middle junctions. — The network. The upper path is $r$ then $R$ , the lower path $R$ then $r$ , and a third $r$ joins their middle junctions.

Given

$r$ , $R$ as placed in the network

Find

R_{AB}

Solution

Step 1A symmetry that swaps the terminals

Name the middle junctions $C$ (upper) and $D$ (lower), hold $A$ at the potential $V$ and $B$ at zero.

The network with the middle junctions named: C on the upper path, D on the lower. — The network with the middle junctions named: $C$ on the upper path, $D$ on the lower.

Turning the drawing through $180^\circ$ about its centre swaps $A$ with $B$ and $C$ with $D$ , and carries every resistor onto one of the same value: $r$ on $AC$ onto $r$ on $DB$ , $R$ on $CB$ onto $R$ on $AD$ , and the rung onto itself. Swapping the terminals reverses the applied voltage, so each potential $\varphi$ goes to $V - \varphi$ at the image point:

\varphi_D = V - \varphi_C \tag{1}

Step 2The current rule at

C

Write $x = \varphi_C$ . The current arriving at $C$ through $r$ from $A$ leaves through $R$ to $B$ and through the rung to $D$ . Using (1) for $\varphi_D$ :

\frac{V - x}{r} = \frac{x}{R} + \frac{x - (V - x)}{r}

Multiplying by $rR$ and collecting the terms in $x$ :

\begin{aligned} R(V - x) &= rx + R(2x - V)\\[2pt] 2RV &= x\,(3R + r) \end{aligned}

\varphi_C = \frac{2RV}{3R + r}

(2)

Step 3The current drawn from

A

Two currents leave $A$ : through $r$ to $C$ and through $R$ to $D$ . By (1), $V - \varphi_D = \varphi_C$ , so the second is $\varphi_C/R$ :

I = \frac{V - \varphi_C}{r} + \frac{\varphi_C}{R}

Substituting (2), with $V - \varphi_C = V(R + r)/(3R + r)$ :

I = \frac{V(R + r)}{r(3R + r)} + \frac{2V}{3R + r} = \frac{V\,(R + r + 2r)}{r(3R + r)} = \frac{V(R + 3r)}{r(3R + r)}

The resistance is $V/I$ :

R_{AB} = \frac{r\,(r + 3R)}{R + 3r}

(3)

Check

Three cases that reduce by hand. With $R = r$ all five resistors are equal, the bridge balances, the rung carries nothing, and two paths of $2r$ in parallel give $r$ ; (3) gives $r \cdot 4r/4r = r$ . As $R \to \infty$ the two $R$ arms are cut, leaving the single path $A$ , $C$ , $D$ , $B$ through three resistors $r$ , that is $3r$ ; the formula tends to $r \cdot 3R/R = 3r$ . As $R \to 0$ , $D$ merges with $A$ and $C$ with $B$ , so all three resistors $r$ stand in parallel between them, giving $r/3$ ; the formula gives $r \cdot r/3r = r/3$ .

Remark

By (1) and (2) the rung voltage is $\varphi_C - \varphi_D = V(R - r)/(3R + r)$ : it vanishes only at $R = r$ and changes sign with $R - r$ , so the rung current runs from $C$ to $D$ when $R > r$ and the other way when $R < r$ . Exchanging $r$ and $R$ everywhere gives the other crossed bridge, of resistance $R(R + 3r)/(r + 3R)$ , and its product with (3) is exactly $rR$ : the two networks are reciprocal about the geometric mean of the resistors.