Two batteries of e.m.f4 V and 8 V with internal resistance of 1 `Omega` and 2 `Omega` are connected in a circuit with a resistance of 9 `Omega` as shown in the figure . The current and potential difference between the point P and Q are

Options

A. `(1)/(3) A and 3V`
B. `(1)/(6) A and 4V`
C. `(1)/(9)A and 9V`
D. `(1)/(12)A and 12V`

Answer

A

Explanation

Given : E.M.F of first battery `(E_(1)) = 4 V ,`
E.M.F of second battery `(E_(2)) = 8 V `,
Internal resistance of first battery `(r_(1)) = 1 Omega` ,
Internal resistance of second battery `(r_(2)) = 2 Omega` and
Circuit resistance (R) ` = 9 Omega`
We know from the Kirchhoff's second law that in the close circuit
PRQP .
`IR+Ir_(2) - E_(2) + E_(1) + Ir_(1) = 0 ` (or)
`(Ixx 9) + (I xx 2) - 8 + 4 + (I xx 1) = 0 `
(or) `9I + 2I - 4 + I = 0 ` (or)
` 12 I = 4 (or) I = (4)/(12) = (1)/(3)` A
`:.` Potential difference across the points P and Q
`(V_(PQ)) = I xx R = (1)/(3) xx 9 = 3 V ` .