question_answer

Two batteries of emf 4 V and 8 V with internal resistance \[1\Omega \]and \[2\Omega \] respectively are connected to an external resistance \[R=9\Omega \] as shown in figure. The current in circuit and the potential difference between P and Q respectively will be

A) \[\frac{1}{2}A,9V\]                         

B) \[\frac{1}{12}A,12V\]

C) \[\frac{1}{3}A,3V\]       

D) \[\frac{1}{6}A,4V\]

Correct Answer: C

Solution :

                  Net emf \[{{E}_{2}}-{{E}_{1}}=8V-4V=4V\] Net resistance \[=R+{{r}_{1}}+{{r}_{2}}\] \[=\text{9}+\text{1}+\text{2}\] \[=12\Omega \] From Ohm's law                                 \[V=iR\] \[i=\frac{V}{R}=\frac{4}{12}=\frac{1}{3}A\] Potential difference between P and \[Q\] = Potential difference across R \[=lR=\frac{1}{3}\times 9=3v\]