Question

In a metre bridge experiment, resistance box (with R = 2Ω) is connected in the left gap and the unknown resistance S in the right gap. If balancing length be 40 cm, calculate value of S. * (a) 2Ω (b) 3 Ω (c) 4 Ω (d) 2.5 Ω​

Answer 1

Answer:

d) 2.5 Ω

Explanation:

In a metre bridge experiment, resistance box (with R = 2Ω) is connected in the left gap and the unknown resistance S in the right gap. If balancing length be 40 cm, calculate value of S. * (a) 2Ω (b) 3 Ω (c) 4 Ω (d) 2.5 Ω

Answer 2

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Case 1 :

Null point is found at

1a=40

Using balanced Wheatstone bridge condition,

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

x

2

=

100−1a

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}

x

′

2

=

100−1b

1b

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

x

′

2

=

100−50

50

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}x=

x+r

xr

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

3+r

3r

⟹ R=6Ω

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

x = 2 \: ohmx=2ohm

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