Question:

Published on: 3 August, 2024

**A Wheatstone bridge has the following resistances : AB = 200Ω, BC = 20Ω, CD = 8Ω & DA = 100Ω. A galvanometer of 40Ω is connected across BD. Find the current through the galvanometer when 20V is applied across A.C.**

Answer:

Resistance of unknown resistor required for balance=\(R_{AD}=\left(\frac{R_{AB}}{R_{BC}}\right)\ast R_{CD}\)

\(=\frac{200}{20}\ast8=80\Omega\)

In the actual bridge unknown resistor has a value of 100Ω

\(\Delta R=(100-80)\Omega=20\Omega\)

Thevenin source generator emf \(E_0=E\left[\frac{R_{AD}}{R_{AD}+R_{CD}}-\frac{R_{AB}}{R_{AB}+R_{BC}}\right]\)

\(=20\left[\frac{100}{100+8}-\frac{200}{200+20}\right]=20\left[0.9259-0.9090\right]=0.33618\)

Internal resistance of bridge looking into terminals B and D.

\(R_0=\frac{R_{AD}R_{CD}}{R_{AD}+R_{CD}}+\frac{R_{AB}R_{BC}}{R_{AB}+R_{BC}}=\frac{100\ast8}{100+8}+\frac{200\ast20}{200+20}=\frac{800}{108}+\frac{4000}{220}\)

\( =7.4074+18.181=25.5892\Omega\)

Current through the galvanometer=\(I_g=\frac{E_0}{R_0+G}=\frac{0.33618}{25.5892+40}=\frac{0.33618}{65.5892}=0.00512A\)

\(=5.12mA \)

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