The resistance of the meter bridge AB in given figure is 4 Ω. With a cell of emf ε = 0.5 V and rheostat resistance Rh = 2 Ω the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε2 the same null point J is found for Rh = 6 Ω. The emf ε2 is:

Engineering

Physics

Wheatstone Bridge and Meter Bridge and Potentiometer

Question

The resistance of the meter bridge AB in given figure is 4 Ω. With a cell of emf ε = 0.5 V and rheostat resistance R_h = 2 Ω the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε₂ the same null point J is found for R_h = 6 Ω. The emf ε₂ is:

0.4

0.5

0.6

0.3

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$0 .5 = \frac{6}{(2 + λL)} λx . ....... (1)$

$E_{2} = \frac{6}{(6 + λL)} λx . ..... (2)$

So dividing equation (1) and (2)

$\frac{E_{2}}{0 .5} = \frac{2 + 4}{6 + 4} = \frac{3}{5}$

⇒ E₂ = 0.3 Volt.