question_answer

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

A)  \[\frac{2E}{4r+R}\]

B)  \[\frac{3E}{4r+R}\]

C)  \[\frac{3E}{3r+R}\]

D)  \[\frac{2E}{3r+R}\]

Correct Answer: A

Solution :

 Total emf of the cell \[=3E-E=2E\] Total internal resistance \[=4r\] \[\therefore \] Total resistance of the circuit \[=4r+R\] So, the current in the external circuit \[\left( \because \,i=\frac{V}{R} \right)\] \[\therefore \] \[i=\frac{2E}{4r+R}\]