question_answer

Two batteries of emf 4 V and 8 V with internal resistances 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) \[\frac{1}{9}A,9V\]         

B)        \[\frac{1}{12}A,12V\]    

C)        \[\frac{1}{3}A,3V\]         

D)        \[\frac{1}{6}A,4V\]

Correct Answer: C

Solution :

In a cell the emf E of the cell (which is responsible for the flow) is directed from the negative electrode to the positive electrode inside the cell. Hence, in the given circuit 8 V cell is supplying the current while the 4V cell is being charged by taking the current. Net \[emf={{E}_{2}}-{{E}_{1}}=8V-4V=4V\] Net resistance \[=R+{{r}_{1}}+{{r}_{2}}\] \[=9+1+2=12\,\Omega \] From Ohms law \[V=iR\] \[\therefore \]  \[i=\frac{V}{R}=\frac{4}{12}=\frac{1}{3}A.\] Potential difference between P and Q = potential difference across R \[=IR=\frac{1}{3}\times 9=3V\]