Questions

# The resistances in the two arms of the meter bridge are 5 $\Omega$ and R $\Omega$ respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6${\mathcal{l}}_{1}$.The resistance R is

## Remember concepts with our Masterclasses.

80k Users
60 mins Expert Faculty Ask Questions
a
10 Ω
b
15 Ω
c
20 Ω
d
25 Ω

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is B

In the first case, At the balance point5R=l1100-l                              . . . . . (i)In the second case At the balance point5(R/2)=1.6l1100-1.6l1            . . . . . (ii)Divide eqn. (i) by eqn. (ii), we get 12=100-1.6l11.6100-l1160-1.6l1=200-3.2l11.6l1=40   or   l1=401.6=25 cmSubstituting this value in eqn. (i), we get 5R=2575R=37525Ω=15Ω

Similar Questions

In a meter bridge circuit as shown, when one more resistance of 100 Ω is connected is parallel with unknown resistance x, then ratio $\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}$become 2. If l1 is balance length of the uniform wire A B, then the value of x must be: