What is end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5 Ω and S respectively. When the resistance S is shunted with an equal resistance, the new balance length found to be 1.5 l1 , where l1 is the initial balancing length. Calculate the value of S.
Answers
End error: The shifting of zero of the scale at different points as well as the stray resistance due to non-uniformity of metal wire gives rise to end error in meter bridge. How to overcome end error? End error can be removed by repeating the experiment by interchanging unknown and known resistance and finally taking the mean value of resistance obtained in each experiment. Given: The resistance in two arms of meter bridge are R = 5 Ω and 5 Ω. According to Wheatstone bridge principle, we have the condition at balancing point as After shunting the resistance S by equal resistance, new balance length is 1.5l1. Shunting means connecting resistance in parallel. Thus, after shunting the new resistance, we get and l1 = 1.5l1. Using Wheatstone bridge principle, we get Using Eq. (1), we get That is, Substituting the value of l1 = 100/3 and R = 5 in Eq. (1)
End error in a meter bridge:
The resistance and stray resistance changes at different points of the scale due to non-uniformity of the wire.
Overcoming end error in a meter bridge:
The end error is overcome by repeating the experiment by changing the known and unknown resistance.
The value of S is 10 ohms.
According to the question
R/S = P/Q = I/(100 - 1)
5/S = I1/(100 - 1) → (1)
(5/S)/2 = 1.5 × I1/(100 - (1.5 × I1))
10/S = 1.5 × I1/(100 - (1.5 × I1)) → (2)
Solving equation (1) and (2) , we get,
I1 = 100/3 cm
∴ S = 10 ohm