In the given Wheatstone bridge, the current in the resistor 3 R is zero. Find the vlaue of R, if the carbon resistor, connected in one arm of the bridge, has the colour sequence of red, red and orange.

The resistance of BC and CD arms are now interchanged and another carbon resistance is connected in place of R so that the current through the arm BD is agian zero. Write the sequence of colours bands of this carbon resistor. Also find the nature of current through it.

Explanation

Let S be the resistance of carbon resistor in arm DC. Its value is `S = 22 xx 10^(3)Omega`.
Since the current in BD arm is zero, so the Wheatstone bridge is balanced. Therefore,
`(2R)/(2R) = R/S`
So, `R = S = 22 xx 10^(3)Omega = 22 K Omega`
When the resistance in arms BC and CD are interchanged and another carbon resistor R' is connected in place of R. The condition for balanced bridge will be
`(2R)/R = (R')/(2R`)
or `R' = 4R = 4 xx 22 k Omega = 88 kOmega`
`= 88 xx 10^(3)Omega`
The sequence of colours will be grey, grey and orange.