CBSE Class 12 · 2009 · All India · Current ElectricityShort Answer

(i) State the principle of a metre bridge. (ii) In a metre bridge the balance point is at distance

ℓ_{1}

with resistances R and S. When an unknown resistance

X

is connected in parallel with S, the balance point shifts to

ℓ_{2}

. Find the expression for

X

in terms of

ℓ_{1}, ℓ_{2}

and S.

Solution

(i) A metre bridge is a balanced Wheatstone bridge; at balance

\frac{R}{S} = \frac{ℓ _{1}}{100 - ℓ _{1}}

. (ii) With

X ∥ S

, the right-gap resistance is

S_{1} = \frac{S X}{S + X}

and

\frac{R}{S _{1}} = \frac{ℓ _{2}}{100 - ℓ _{2}}

. Dividing this by the first relation and simplifying,

\frac{S}{X} = \frac{100 ( ℓ _{2} - ℓ _{1} )}{ℓ _{1} ( 100 - ℓ _{2} )}

, hence

X = \frac{ℓ _{1} ( 100 - ℓ _{2} )}{100 ( ℓ _{2} - ℓ _{1} )} S .

CBSE Class 12 Board · 2009 · All India · Subjective