Question

In a meter bridge with r and s in the gaps the null point is found at 40cm from

a. If a resistance of 30ohm is connected in parallel with s,the null point occurs at 50cm from a determine value of r and s

Answer 1

Answer:

The value of r = 10 Ω and s = 15 Ω.

Step-by-step explanation:

Case 1:  

In the mether bridge with r & s in the gaps the null point is given to be,

l1 = 40 cm  

Therefore, we can write

r/s = l1/ (100-l1)  

⇒ r/s = 40/(100-40)  

⇒ r/s = 40/60 = 2/3

⇒ s = (3/2) * r ……. (i)

Case 2:

A resistance of 30Ω is connected in // with s.

So, we can have

1/s’ = 1/30 + 1/s

⇒ s’ = 30s / (s+30) ….. (ii)

Also, in this case the null point occurs to be at, l2 = 50 cm

∴ r/s’ = l2 / (100-l2)

⇒ r/s’ = 50 / (100-50) = 1

⇒ r = s’ …. (iii)

Case 3:

Substituting the value of s from (i) in (ii), we get

s’ = $\frac{30 * \frac{3r}{2} }{30 + \frac{3r}{2}}$

⇒ s’ = 45r / [30 + (3r/2)]

from (iii) we know r = s’, so

⇒ r = 45r / [30 + (3r/2)]

⇒ 45 = [30 + (3r/2)]

⇒ 3r/2 = 15

⇒ r = 10 Ω

and,

s = 3/2 * r = 3 * 10 /2 = 15 Ω