Question

In a meter bridge when the resistance in the left gap is 4 Omega and an unknown resistance in the right gap the balance point is obtained at 40 cm from the zero end. On shunting the unknown resistance with 4 Omega, find the shift of the balance point on the bridge wire

Answer 1

Answer:

22.5...

Explanation:

In the first case , resistance in the left gap `X = 4 Omega` and Y is the resistance in the right gap of meter bridge

`(X )/(Y) = (40)/((100 - 40)) = (40)/(60) = (23)/(3)`

`Y = (3)/(2) X = (2)/(2) xx 4 = 6 Omega`

when `Y` is shouted by `4 Omega` the effective resistance

`Y = (6 xx 4)/(6+4) = 24 Omega`

In second case let the balance point be obtained at length `l ` from the left end of meter bright .The `(4)/(24)= (l)/(100 - l) or (5)/(3) = (l)/(100 - l) or l = 62.5 cm`

She it in the balance point `62.5 - 40 = 22.5 cm`

Answer 2

Balance point is shifted by 22.5 cm

Explanation:

• In wheat stone bridge a wire of 100 cm of uniform cross-section is stretched over a wooden block with a scale which reads up to 1 m then we connect two resistances and get a balance point where galvanometer connected through jockey gives null deflection and get   $\frac{P}{Q} =\frac{R}{S}$

First case :-

• balance point = 40 cm = R
• another point = 100 - 40 = 60 cm = S
• left gap resistance = 4 Ω

• let unknown resistance = x Ω

=>     $\frac{4}{x} = \frac{40}{60}$

=> x = 6 Ω

Second case :-

• shunt with resistance = 4 Ω   (parallel connected with 6 Ω)
• effective resistance $\frac{1}{R}$ = $\frac{1}{R_{1} } +\frac{1}{R_{2} }$

=>      $\frac{1}{4} +\frac{1}{6}$ = $\frac{12}{5}$ Ω

let the new balance point is 'l'

=> another point = (100 - l) cm

   by using following equation

$\frac{P}{Q} =\frac{R}{S}$

=> $\frac{4}{\frac{12}{5} } =\frac{l}{100-l}$

=> 2000 - 20 l =12 l

=> l = 62.5 cm

Shift in the balance point = 62.5 - 40 = 22.5 cm

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