
In a meter bridge shown in the figure, the balance point is found to be 40cm from and A. If a resistance of $10\Omega $is connected in series with R, a balance point is obtained 60cm from A. Calculate the value of R and S.
Answer
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Hint: Meter bridge: A meter bridge also called a slide wire 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.
Complete step-by-step solution:
Given,
R=$10\Omega $
When a balance point is at 40cm, we have
$ \Rightarrow \dfrac{R}{S} = \dfrac{{40}}{{100 - 40}}$
Simplify,
$ \Rightarrow \dfrac{R}{S} = \dfrac{{40}}{{60}}$
$ \Rightarrow \dfrac{R}{S} = \dfrac{2}{3}$
$ \Rightarrow 3R = 2S$ …(1)
When a resistance of $10\Omega $is added in series with R, then equivalent resistance at R is${R^`} = R + 10$
Now, balance point is at 60 cm. So, we have
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{{60}}{{100 - 60}}$
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{{60}}{{40}}$
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{3}{2}$
$ \Rightarrow 2R + 20 = 3S$
$ \Rightarrow 2R - 3S = 20$ …(2)
Solving equation 1 and 2
S=$12\Omega $
R=$8\Omega $
Note: The device required in finding the unknown resistance of a conductor using the meter bridge are- meter bridge, resistance box, galvanometer, unknown resistance of length 1m, screw gauge, connecting wire, jockey, and one way key. A scale is attached to the block.
Complete step-by-step solution:
Given,
R=$10\Omega $
When a balance point is at 40cm, we have
$ \Rightarrow \dfrac{R}{S} = \dfrac{{40}}{{100 - 40}}$
Simplify,
$ \Rightarrow \dfrac{R}{S} = \dfrac{{40}}{{60}}$
$ \Rightarrow \dfrac{R}{S} = \dfrac{2}{3}$
$ \Rightarrow 3R = 2S$ …(1)
When a resistance of $10\Omega $is added in series with R, then equivalent resistance at R is${R^`} = R + 10$
Now, balance point is at 60 cm. So, we have
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{{60}}{{100 - 60}}$
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{{60}}{{40}}$
$ \Rightarrow \dfrac{{R + 10}}{S} = \dfrac{3}{2}$
$ \Rightarrow 2R + 20 = 3S$
$ \Rightarrow 2R - 3S = 20$ …(2)
Solving equation 1 and 2
S=$12\Omega $
R=$8\Omega $
Note: The device required in finding the unknown resistance of a conductor using the meter bridge are- meter bridge, resistance box, galvanometer, unknown resistance of length 1m, screw gauge, connecting wire, jockey, and one way key. A scale is attached to the block.
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