In the circuit shown, if a conducting wire is connected between point `A` and `B`, the current in this wire will

Options

A. flow from `A` to `B`
B. flow in the direction which will be decided by the value of `V`
C. be zero
D. flow from `B` and `A`

Answer

D

Explanation

(d) Current will flow from higher to lower potential.
Resistance `4 Omega` and `4 Omega` are connected in series, so their effective resistance, `R' = 4 + 4 = 8 Omega`
Similarly, `1 Omega` and `3 Omega` are in series, `R" = 1 + 3 = 4 Omega`
Now `R'` and `R"` will be in parallel, hence effective resistance `R = (R' xx R'')/(R' + R'') = (8 xx 4)/(8 + 4) = (32)/(12) = (8)/(3) Omega`
Current through the circuit, from Ohm's law
`i= (V)/(R ) = (3V)/(8) A`
Let current `i_(1)` and `i_(2)` flow in the branches as shown
`8 i_(1) = 4 i_(2)`
`i_(2) = 2 i_(1)` .....(i)
Also `i= i_(1) + 2 i_(1)`
`i_(1) = (V)/(8) A` and `i_(2) = (V)/(4) A`
Potential drop at `A, V_(A) = 4 xx i_(1) = (4 V)/(8) = (V)/(2)`
Potential drop at `B, V_(B) = 1 xx i_(2) = 1 xx (V)/(4) = (V)/(4)`
Since, drop of potential is greater in `4 Omega` resistance. It will be at lower potential than `B`, hence, on connecting wire between points `A` and `B`, the current will flow from `B` to `A`.