Question

A battery of emf `epsilon_0 = 5V` and internal resistance `5 Omega` is connected across a long uniform wire AB of length 1m and resistance per unit length `5 Omegam^(-1)`. Two cells of `epsilon_1 = 1 V and epsilon_2 = 2V` are connected as shown in the figure.

A. the null point is at A.
B. If the jockey is touched to point B, the current in the galvanometer will be going towards B.
C. When jockey is connected to point A, no current flows through 1 V battery.
D. The null point is at distance of `8//15` m from `A_2`.

Answer 1

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.