Two batteries of e.m.f. `4V` and `8V` with internal resistances `1 Omega` and `2 Omega` are connected in a circuit with a resistance of `9 Omega` as shown in figure. The current and potential difference between the points `P` and `Q`

Options

A. `(1)/(3)A and 4 V`
B. `(1)/(3) A and 3 V`
C. `(1)/(2)A and 5 V`
D. `(1)/(6)A and 3 V`

Answer

B

Explanation

`therefore 8 V gt 4 V`, the current will flow in the circuit from `E_(1) " to " E_(2)`
Let i be the current flowing in the circuit.
Applying Kirchhoff's voltage law to the loop RSQP, we get
`-2i+8-4-1i-9i=0`
` therefore -12i = -4 therefore i=(4)/(12)=(1)/(3) A`
`therefore` P.D. between P and Q=P.D. across the `9 Omega` resistance
`=iR=(1)/(3)=xx9=3 V`
[Thus `i=(1)/(3)A and V_(PQ)=3V]`