Question:
Published on: 10 August, 2022

A Wheatstone bridge consists of AB = 4 Ω, BC = 3 Ω, CD = 6 Ω and DA = 5 Ω, A 2.4 V battery is connected between points B and D. A galvanometer of 8 Ω resistances is connected between A and C. Using Thevenin’s theorems find the current through the galvanometer.

Fig.28(a)

Total resistance of parallel pair $$\frac{9 \times 9}{ 9 + 9}=4.5\ \mathrm{\Omega}$$

Source current is $$\frac{2.4}{4.5}=0.53\ Amp$$.

Current the branch BAD is $$\frac{2.4}{4.5 \times 2} = \frac{1.2}{4.5}=0.27\ Amp$$.

Current through the branch BCD is $$\frac{2.4}{4.5 \times 2}=\frac{1.2}{4.5}=0.27\ Amp$$.

Voltage drop across BA is $$\frac{4 \times 1.2}{4.5}=1.07\ V$$

Voltage drop across BC is $$\frac{3 \times 1.2}{4.5}=0.8\ V$$

Fig. 28(b)

Fig. 28(c)

Hence voltage across AC is $$V_{o.c}=1.07-0.8=0.27\ V$$

To find the Rth the circuit diagram may be redrawn as

Hence Rth =$$\left(8+2.2+2\right)=12.2\ \mathrm{\Omega}$$

The current through the galvanometer is $$\frac{0.27}{12.2}=0.02\ Amp$$.

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