question_answer

Two batteries of emf 4 V and 8 V with internal resistance 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 are

A)  \[\frac{1}{3}A\]and 3 V                

B)  \[\frac{1}{6}A\] and 4V

C)  \[\frac{1}{9}A\]and 9 V                

D)  \[\frac{1}{2}A\]and 12 V

Correct Answer: A

Solution :

                Applying Kirchhoffs voltage law in the given loop \[\Rightarrow \]               \[i=\frac{1}{3}A\] Potential difference across\[PQ=\frac{1}{3}\times 9=3V\]