The resistances in the two arms of the meter bridge are 5 ? and R? respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6. The resistance R is [2014]

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Solution

Q.
The resistances in the two arms of the meter bridge are 5 Ω and RΩ respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 $l_{1}$ . The resistance R is [2014]

1 10 Ω

2 15 Ω

3 20 Ω

4 25 Ω

Ans.
(2)

In the first case,

At balance point

$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$ ...(i)

In the second case,

At balance point

$\frac{5}{(R / 2)} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$ ..(ii)

Divide eqn. (i) by eqn. (ii), we get

   $\frac{1}{2} = \frac{100 - 1.6 l_{1}}{1.6 (100 - l_{1})}$

or    $160 - 1.6 l_{1} = 200 - 3.2 l_{1}$

   $1.6 l_{1} = 40$ or   $l_{1} = \frac{40}{1.6} = 25 cm$

Substituting this value in eqn. (i), we get:

   $\frac{5}{R} = \frac{25}{75}$ or   $R = \frac{375}{25} Ω = 15$ $Ω$

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