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\[3V\]                  

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 :

\[{{E}_{1}}=4V\]and \[{{E}_{2}}=8\,V\] So, \[{{E}_{2}}>{{E}_{1}}\]current flows from Q to P. \[\therefore \]      \[i=\frac{1}{3}A\] \[\therefore \] Potential difference actors \[PQ=\frac{1}{3}\times 9=3\,V\] Hence, the correction option is [a].