A 5V battery with internal resistance `2 Omega` and a 2V battery with internal resistance `1 Omega` are connected to a `10 Omega` resistor as shown in the figure. The current in the `10 Omega` resistor is

Options

A. 0.27A,`P_2 to P_1`
B. 0.03A `P_1 to P_2`
C. 0.03A `P_2 to P_1`
D. 0.27A `P_1 to P_2`

Explanation

Let potential at `P_1` be 0V and potential at `P_2` be `V_0` Now apply KCL at `P_2`
`sum I=0`
`(V_0-5)/2+(V_0-0)/10+(V_0-(-2))/1 =0 or V_0=5/16`
So current through `10 Omega` resistor is `V_0/10` from `P_2` to `P_1`