Balance the metre bridge carefully

Electricity and Magnetism Level 2

A resistance of 2 Ω is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistance are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance ( in Ω ) is

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Sayan Chaudhuri by sayan chaudhuri , 25, India

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2 solutions

Crazy Singh
Dec 23, 2013

lets say the unknown resistance is R R . so 2 R = x 100 − x \frac{2}{R} = \frac{x}{100-x} .

in the second step the resistances are interchanged. so R 2 = y 100 − y \frac{R}{2} = \frac{y}{100-y} .

and it is given that y − x = 20 y-x = 20 . solving above equations we get R = 3 R = 3

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yeah......... a short nd sweet solution

Mayankk Bhagat - 7 years, 5 months ago

If the balancing length is measured initially on the side of the unknown resistance X, it will shift from 60 cm to 40 cm. [Remember that the balancing length is measured from the same side before and after interchanging. You might have noted that the balance point in a meter bridge shifts symmetrically with respect to the mid point of the bridge wire].

Therefore, we have X/2 = 60/40, from which X = 3Ω.

[If the balancing length is measured on the side of the known resistance, it will change from 40 cm to 60 cm. In this case, 2/X = 40/60, from which X =3Ω].

Aman Kumar - 7 years, 4 months ago

If the balancing length is measured initially on the side of the unknown resistance X, it will shift from 60 cm to 40 cm. [Remember that the balancing length is measured from the same side before and after interchanging. You might have noted that the balance point in a meter bridge shifts symmetrically with respect to the mid point of the bridge wire].

Therefore, we have X/2 = 60/40, from which X = 3Ω.

[If the balancing length is measured on the side of the known resistance, it will change from 40 cm to 60 cm. In this case, 2/X = 40/60, from which X =3Ω].

Muhammad Sami - 7 years, 3 months ago

I made the correct solution but select the wrong solution by wrong :\ :\

Moon Walker - 7 years, 4 months ago
Mayankk Bhagat
Jan 7, 2014

let "x" be the unknown resistance & "t" be the length of its corresponding wire portion. then, we have
x/2=t/(100-t).........(I) and 2/x=t-20/(120-t)...(2).......... on solving 1 and 2 we get- t=60.......... put t=60 in either equation 1 or2 . and we shall get x=3. simple!!!!!!!!!!!!!

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