Question

In a metre bridge experiment , resistance box with 2 Ω is connected in the left gap and the unknown resistance S in the right gap. If the balancing length be 40 cm, value of S will be *

(a) 2 Ω

(b) 3 Ω

(c) 4 Ω

(d) 5 Ω



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Answer 1

Question

A metre bridge experiment , resistance box with 2 Ω is connected in the left gap and the unknown resistance S in the right gap. If the balancing length be 40 cm, value of S will be *

Answer

Case 1 :

Null point is found at

1a=40

Using balanced Wheatstone bridge condition,

$\frac{2}{x} = \frac{1a}{100 - 1a}$

⟹ X=3Ω

Case 2 :

New null point is found at

1B =50 cm

Using balanced Wheatstone bridge condition,

⟹ X' =2Ω

Since the unknown resistance gets decreased, thus we have to connect a resistance R in parallel to X so that X' comes out to be 2Ω.

Thus resistance of parallel combination

$\frac{2}{x'} = \frac{1b}{100 - 1b}$

$\frac{2}{x'} = \frac{50}{100 - 50}$

$x = 2 \: ohm$

Since the unknown resistance gets decreased, thus we have to connect a resistance R in parallel to X so that X′

comes out to be 2Ω.

Thus resistance of parallel combination

$x = \frac{xr}{x + r}$

$2 = \frac{3r}{3 + r}$

⟹ R=6Ω

Hence 6Ω must be connected in parallel to unknown resistance X.

$bts ❤ exo$

Answer 2

Given: In a metre bridge experiment, resistance box with 2Ω is connected in the left gap and the unknown resistance s in the right gap.

To find: Value of s

Explanation: In a metre bridge experiment, the resistance of an unknown resistance can be found with the help of a known resistance.

At the balancing length no current flows in the wire and the potential is balanced.

The formula for knowing the unknown resistance is:

=>$\frac{r}{s} = \frac{l}{100 - l}$

=>$\frac{2}{s} = \frac{40}{60}$

=>$\frac{2}{s} = \frac{2}{3}$

=>s = 3

Therefore, the value of s is 3 ohm.