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Question: In a metre bridge the null point is found at a distance of 40cm from \(A\). If a resistance of \(12\...

In a metre bridge the null point is found at a distance of 40cm from AA. If a resistance of 12Ω12\Omega is connected in parallel with SS, the null point occurs at 50cm50cm from AA. Determine value of RR and SS.

Explanation

Solution

Here two conditions are given, in one condition a resistance of 12Ω12\Omega is connected in parallel to SS and in another condition there is no such resistance connected in parallel. Applying the condition of Wheatstone bridge in these two conditions will give two different linear equations in form of variables of RR and SS . After solving the two equations you will get the answer.

Complete step by step solution:
Here in this question two conditions are given,
First, when a resistance of 12Ω12\Omega is not connected in parallel to SS . In that case we can write,
RS=l1(100l1)\dfrac{R}{S} = \dfrac{{{l_1}}}{{(100 - {l_1})}}
Putting l1=40cm{l_1} = 40cm as given in the question we have,
RS=4010040\dfrac{R}{S} = \dfrac{{40}}{{100 - 40}}
So the relation between RR and SS is given by,
R=23SR = \dfrac{2}{3}S

Now case second when 12Ω12\Omega resistance is connected in parallel to the resistor SS .
In this case the effective resistance can be written as,
1S1=1S+112\dfrac{1}{{{S_1}}} = \dfrac{1}{S} + \dfrac{1}{{12}}
On simplifying this expression we have,
1S1=12+512S\dfrac{1}{{{S_1}}} = \dfrac{{12 + 5}}{{12S}}
Taking reciprocals on both sides we have,
S1=12S(12+5){S_1} = \dfrac{{12S}}{{(12 + 5)}}
Now writing the condition of Wheatstone bridge we have,
RS1=l100l\dfrac{R}{{{S_1}}} = \dfrac{{{l'}}}{{100 - {l'}}}
Putting the expression for S1{S_1} we have,
R=12S(12+S)×5050R = \dfrac{{12S}}{{(12 + S)}} \times \dfrac{{50}}{{50}}
On simplifying the above expression we have,
R=12S12+SR = \dfrac{{12S}}{{12 + S}}
Now we have two expressions for RR after equating them we have,
23S=12S(12+S)\dfrac{2}{3}S = \dfrac{{12S}}{{(12 + S)}}
On simplifying the above expression we have,
12+S=1812 + S = 18
So we have, S=6ΩS = 6\Omega
So we have R=23×6=4ΩR = \dfrac{2}{3} \times 6 = 4\Omega

So, the values of RR and SS are 4Ω4\Omega and 6Ω6\Omega respectively.

Note: It is important to note the working principle of a meter bridge. A meter bridge is an instrument that works on the principle of a Wheatstone bridge. A meter bridge is used in finding the unknown resistance of a conductor as that of in a Wheatstone bridge. The null point of a Wheatstone is also known as the balance point of the Wheatstone bridge.