  Question:
Published on: 10 August, 2022

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.

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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