Question

Six lead-acid type of secondary cells each of EMF 3 V and internal resistance 0.30 $\Omega$ are joined in series in the same direction to provide a supply to a resistance of 10 $\Omega$. What is the potential drop across the external resistor?

• A. 13.2 V
• B. 15.2V
• C. 17.2V
• D. 19.2V

Answer 1

The correct option is B 15.2V

We have 6 cells in the same direction. So their EMF’s and internal resistances get added up.
So find the total EMF and R of the combination.
E=6 $\times$ 3 = 18V
R=6 $\times$ 0.3 = 1.8 $\Omega$
R’ = 10 $\Omega$
Since all these are in series, we just have to find current.
$I=\frac{E}{R+R^{\prime }}$
I=1.52A
So potential drop across 10 $\Delta$ is
V=IR’ = 15.2V