Question

In a Wheatstone bridge circuit
P=5ohm
Q=6ohm
R=10ohm
S=5ohm
Find the additional resistance to be used in series with S so that the bridge is balanced

Answer 1

Answer:

7 ohm

Explanation:

p/q=r/s is the common relation

let new resistance be x to be used in series

to balance the S resistance

then new relation

p/q=r/s+x

5/6=10/5+x

5(5+x)=10×6

5(5+x)=60

25+5x=60

5x=35

x=7ohm

Answer 2

Answer:

7 ohms

Explanation:

Let the additional resistance to be used = x

'x' is connected in series with S.

S + x = Q/P X R

5 + x = 6/5 X 10 = 6 X 2 = 12

5 + x = 12

x = 12 - 5

x = 7 ohms