Question

In a practical Wheatstone bridge circuit, wire AB is 2 m long. When resistance Y = 2.0W and jockey is in position J such that AJ = 1.20 m, there is no current in galvanometer, find the value of unknown resistance X. The resistance per unit length of wire AB = 0.01 W/cm. Also calculate the current drawn by the cell of emf 4.0 V and negligible internal resistance..

Answer 1

P = Resistance of wire AJ

= (1.20 × 100 cm) × (0.01Ω/cm) = 1.20 Ω

Q = Resistance of wire BJ

= [(2 – 1.20) m × 100] × resistance per cm

= 0.80 × 100 cm × 0.01 Ω = 0.80Ω

Y = 2.0 Ω, X = ?

When no current flows through the galvanometer, the bridge is balanced so

\(\frac{P}{Q}=\frac{X}{Y}\) \(\Rightarrow X=\frac{P}{Q}Y\) 

or \(X=\frac{1.20}{0.80}\times2.0=3.0\Omega\) 

Total resistance of X and Y connected in series

R₁ = X + Y = 3.0 + 2.0 = 5.0 Ω

Total resistance of P and Q connected in series (or wire AB)

R₂ = 2 × 100 × 0.01 = 2.0Ω

The resistance R₁ and R₂ are in parallel, so effective resistance between terminals A and B of bridge is

\(R_{AB}=\frac{R_1R_2}{R_1+R_2}=\frac{5.0\times2.0}{5.0+2.0}=\frac{10}{7}\Omega\)

Current drawn from battery \(I=\frac{\epsilon}{R_{AB}}=\frac{4.0}{10/7}=2.8A\)