question_answer

Two batteries of emf 4 V and 8 V 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 are

A)  \[\frac{1}{3}A\,\,and\,\,3V\]    

B)  \[\frac{1}{6}A\,\,and\,\,4V\]

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

D)  \[\frac{1}{2}A\,\,and\,\,12V\]

Correct Answer: A

Solution :

                 Applying Kirchhoffs law in the given loop                 \[-2I+8-4-1\times I-9I=0\] \[\Rightarrow \]               \[I=\frac{1}{3}A\] Potential difference across PQ                 \[=\frac{1}{3}\times 9=3\,V\]