In the given Wheatstone bridge, the current 3R is zero. Find the value of R, if carbon resistor, connected in one arm of the bridge has the colour sequence of red, red and, orange.
The resistance of BC and CD arm are now interchanged and another carbon resistor is connected in place of R so that the current through arm BD is again zero. Write the sequence of colour bands of the carbon resistor. Also find the current through it.
For no current through BD, the Wheatstone bridge is balanced and the resistance of carbon resistor is R’ = R. = 22000 Ω.
When, resistance of BC and CD arms are interchanged and another carbon resistor is connected in place of R, the current through BD is again zero.
Therefore, again for balanced Wheatstone bridge
R’ = 4R
= 4 x 22000 Ω
= 88000 Ω
For 88000 Ω the sequence of colour bands is gray, gray and orange.
Now, net resistance of arm ADC,
= 88000 + 44000
= 132000 Ω
∴ current through carbon resistor,
