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

The current in the `10Omega` resistor is

Options

A. `0.27A P_2 to P_1`
B. `0.03A P_1 to P_2`
C. `0.03AP_2to P_1`
D. `0.27 A P_1 to P_2`

Answer

C

Explanation

(c) Applying Kirchoff's loop law in `AB P_2P_1A` we get
`-2i + 5 - 10 i_1 = 0` ………(i)

Again applying kirchoff's loop law in `P_2CDP_1P_2` we
get, `10i_1 + 2- I + i_1 = 0` ……….(ii)
From (i) and (ii) `11i_1 + 2 - [(5-10i_1)/2] = 0`
`rArr i_1 = 1/32 A from P_2 to P_1`.