question_answer

Two batteries of emfs 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)  \[\frac{\text{1}}{\text{2}}\text{A}\,\text{and}\,\text{3}\,\text{V}\]     

B)         \[\frac{\text{1}}{6}\text{A}\,\text{and}\,4\,\text{V}\]

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

D)         \[\frac{\text{1}}{2}\text{A}\,\text{and}\,12\,\text{V}\]

Correct Answer: A

Solution :

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