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 \[=lR=\frac{1}{3}\times 9=3v\]