Question

A cell with internal resistance 2.5 ohm is balanced across 450 cm length of potentiometer wire. The cell shunted by a resistance of
4 ohm . calculate the balancing length of wire​

Answer 1

Answer:

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Explanation:

Let balancing length be l

1

and internal resistance be 'r'.

r=R(

l

2

l

1

−l

2

)

r=8(

200

l

1

−200

) ....(1)

r=4(

160

l

1

−160

) ....(2)

From equations (1) and (2)

8(

200

l

1

−200

)=4(

160

l

1

−160

)

10

l

1

−200

=

16

l

1

−160

16l

1

=3200+10l

1

−1600

l

1

=266.66cm

r=8(

200

266.66−200

)

r=2.66Ω

Answer 2

The balancing length of the wire is 731.25 cm

Explanation:

Given:

Internal resistance r = 2.5 ohm

Resistance R = 4 ohm

To find out:

Balancing length of the wire

Solution:

Let the balancing length of the wire is $l_1$

$r=R(\frac{l_1}{l_2}-1)$

$\implies 2.5=4(\frac{l_1}{450}-1)$

$\implies \frac{l_1}{450}-1=\frac{25}{40}$

$\implies \frac{l_1}{450}=\frac{5}{8}+1$

$\implies \frac{l_1}{450}=\frac{13}{8}$

$\implies l_1=\frac{13}{8}\times 450$

$\implies l_1=731.25$ cm

Hope this answer is helpful.

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