Correct Answer - A::B
a., b.
For null point, current flows in the
loop CD only.
`i = (3V)/(2Omega + 1Omega) = 1A`
`V_(CD) = 1V - 1(1) = 0`
Therefore, option (a) is correct. That
is, `V_(A) gt V_(B)`. When jockey touches B, current from A to B increases the
PD across the secondary circuit.
Therefore, option(b) is correct.