Question 533: Current Electricity

Question

When the two known resistance ' $R$ ' and ' $S$ ' are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ' $l_1$ ' from the zero end of a meter bridge wire. An unknown resistance ' X ' is now connected in parallel with ' S ' and null point is found at a distance ' $l_2$ ' form zero end of meter bridge wire. The unknown resistance ' X ' is

Accepted Answer

To determine the unknown resistance $ X $ using a meter bridge, consider the following conditions:
First Condition
Initially, when the resistances $ R $ and $ S $ are in the left and right gaps, the null point is at a distance $ l_1 $ from the zero end. Using the balancing condition of the meter bridge, we have:
$ \frac{R}{l_1} = \frac{S}{100 - l_1} $
Second Condition
When an unknown resistance $ X $ is connected in parallel with $ S $ and the null point shifts to a distance $ l_2 $ from the zero end, the balancing condition becomes:
$ \frac{R}{l_2} = \frac{XS}{(X + S)(100 - l_2)} $
Equating the two conditions, we can derive the relationship:
$ \frac{l_1 S}{100 - l_1} = \frac{l_2 XS}{(X + S)(100 - l_2)} $
Simplifying further:
$ \frac{100 - l_1}{l_1} = \left(1 + \frac{S}{X}\right) \frac{100 - l_2}{l_2} $
Rearranging, we find:
$ 1 + \frac{S}{X} = \frac{l_2 (100 - l_1)}{l_1 (100 - l_2)} $
Subtracting 1 from both sides:
$ \frac{S}{X} = \frac{l_2 (100 - l_1) - l_1 (100 - l_2)}{l_1 (100 - l_2)} $
Which simplifies to:
$ \frac{S}{X} = \frac{100(l_2 - l_1)}{l_1 (100 - l_2)} $
Thus, solving for $ X $:
$ X = \frac{S l_1 (100 - l_2)}{100 (l_2 - l_1)} $

Answer

$\frac{\mathrm{S} l_1\left(100-l_2\right)}{100\left(l_2-l_1\right)}$

Answer

$\frac{\mathrm{S} l_2\left(100-l_1\right)}{100\left(l_1-l_2\right)}$

Answer

$\frac{100\left(l_2-l_1\right)}{\mathrm{Sl}_1\left(100-l_2\right)}$

Answer

$\frac{100\left(l_2-l_1\right)}{\mathrm{S} l_2\left(100-l_1\right)}$

Current Electricity

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