Problem — Wheatstone bridge, unknown resistance

Problem

In a Wheatstone bridge the known arms are $R_1 = 100\ \Omega$, $R_2 = 200\ \Omega$, $R_3 = 300\ \Omega$; the galvanometer reads zero when the fourth arm contains the unknown resistance $R_x$. Determine $R_x$ and explain why the value found does not depend on the bridge’s supply voltage.

Diagram: four-arm bridge $R_1,R_2$ (top), $R_3,R_x$ (bottom); galvanometer G on the diagonal, generator $\mathcal{E}$ on the other.

Solution

Balance condition. The galvanometer reads zero when the two central nodes are at the same potential: no current flows through the galvanometer branch. At balance the two dividers (top branch $R_2$–$R_1$ and bottom branch $R_3$–$R_x$) have the same ratio, i.e. the products of opposite arms are equal: $R_1 R_3 = R_2 R_x \qquad\Longleftrightarrow\qquad \frac{R_2}{R_1} = \frac{R_3}{R_x}$

Solving for the unknown: $R_x = \frac{R_1 R_3}{R_2} = \frac{100 \cdot 300}{200} = 150\ \Omega$

Why it does not depend on the voltage. At balance only the equality of the ratios between the resistances matters: the current in the galvanometer is zero for any value of $\mathcal{E}$. Changing the supply voltage scales all the currents in the arms by the same factor, but leaves the condition $R_1 R_x = R_2 R_3$ unchanged. The bridge is therefore a comparison measurement method, insensitive to the stability of the supply.

$\ev{R_x = 150\ \Omega}$

Links

Topics: Electric circuits Concepts: Resistors in series and parallel