Two cells of same emf E but internal resistance r_{1} and r_{2} are connected in series to an external resistor R (Figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.


Answers (1)

 

EMF_{ef}=E+E=2E

R_{ef}=r_1+r_2

I=\frac{2E}{R+r_1+r_2}

As net potential across the cell is 0,

E-Ir_1=0

E-\frac{2Er_1}{R+r_1+r_2}=0

R=r_1-r_2

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