First slide

meter bridge

Question

The resistances in the two arms of the meter bridge are 5 $Ω$ and R $Ω$ 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

Moderate

A

10 $Ω$
B

15 $Ω$
C

20 $Ω$
D

25 $Ω$

Solution

In the first case,
At the balance point
$\frac{5}{R} = \frac{l_{1}}{100 - l}$ . . . . . (i)
In the second case
At the 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
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{1.6 (100 - l_{1})}$
$160 - 1.6 l_{1} = 200 - 3.2 l_{1}$
$1.6 l_{1} = 40 or l_{1} = \frac{40}{1.6} = 25 cm$
Substituting this value in eqn. (i), we get
$\frac{5}{R} = \frac{25}{75}$
$R = \frac{375}{25} Ω = 15 Ω$

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