Worked exercise — measurement with the Wheatstone bridge

Problem

In a Wheatstone bridge $R_1 = R_2 = 100\;\Omega$ and $R_3 = 47\;\Omega$. Find the value $R_4$ that brings the bridge to balance and the current delivered by the generator, supplied at $\mathcal{E} = 6\;\text{V}$.

Diamond-shaped bridge: at balance the current in the galvanometer $G$ is zero.

Solution

Balance condition. At balance $I_G = 0$ and $R_1/R_2 = R_3/R_4$ holds. Since $R_1 = R_2$, the ratio $R_1/R_2 = 1$, hence: $R_4 = R_3\,\frac{R_2}{R_1} = R_3 = 47\;\Omega$

Generator current. With $I_G = 0$ the central branch is inactive and two parallel paths remain, each formed by two resistors in series: $R_1 + R_3 = 147\;\Omega \qquad R_2 + R_4 = 147\;\Omega$ Being equal, they are two identical resistors in parallel: $R_{\text{eq}} = \frac{147}{2} = 73{,}5\;\Omega \qquad i = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{6}{73{,}5} \approx 0{,}082\;\text{A}$

$\ev{R_4 = 47\;\Omega,\quad i \approx 0{,}082\;\text{A}}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Resistenze in serie e parallelo Competenze: Semplificazione di reti