Problem — balanced Wheatstone bridge

Problem

In a Wheatstone bridge the known arms are $R_1 = 100\ \Omega$, $R_2 = 200\ \Omega$, $R_3 = 150\ \Omega$. The galvanometer between the intermediate nodes measures zero current when $R_x$ takes a certain value. Determine $R_x$ and recall why this method is particularly accurate.

Wheatstone bridge: $G$ is zero when the arms are balanced.

Solution

Balance condition — the galvanometer measures zero current when the two intermediate nodes (to the left and right of $G$) are at the same potential. This happens when the two voltage dividers — $R_1$–$R_3$ on one side, $R_2$–$R_x$ on the other — divide the voltage in the same proportion: $\frac{R_1}{R_3} = \frac{R_2}{R_x}$

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

Why it is accurate — this is a null method: no value is read off a graduated scale (subject to calibration errors of the instrument); instead, the condition where the current in the galvanometer is exactly zero is sought. Under that condition, the result depends only on the ratios between known resistances, not on the sensitivity or linearity of the galvanometer nor on the precise value of the e.m.f. $\mathcal{E}$. It is enough for the galvanometer to be sensitive enough to detect zero; the uncertainty shifts entirely onto the reference resistors, which can be made very accurate.

$\ev{R_x = 300\ \Omega}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Resistenze in serie e parallelo