$(a)$ Six lead-acid type of secondary cells each of $emf$ $2.0 \;V$ and internal resistance $0.015 \;\Omega$ are joined in series to provide a supply to a resistance of $8.5\; \Omega$. What are the current drawn from the supply and its terminal voltage?
$(b)$ $A$ secondary cell after long use has an $emf$ of $1.9 \;V$ and a large internal resistance of $380\; \Omega$. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

(A) Number of secondary cells,$n = 6$.
$Emf$ of each secondary cell,$E = 2.0 \; V$.
Internal resistance of each cell,$r = 0.015 \; \Omega$.
Resistance of the external resistor,$R = 8.5 \; \Omega$.
Since the cells are in series,the total $emf$ is $nE$ and the total internal resistance is $nr$.
The current $I$ drawn from the supply is given by $I = \frac{nE}{R + nr}$.
$I = \frac{6 \times 2.0}{8.5 + 6 \times 0.015} = \frac{12}{8.5 + 0.09} = \frac{12}{8.59} \approx 1.39 \; A$.
The terminal voltage $V$ is given by $V = IR = 1.39 \times 8.5 \approx 11.87 \; V$.
$(b)$ $Emf$ of the cell,$E = 1.9 \; V$.
Internal resistance,$r = 380 \; \Omega$.
The maximum current $I_{max}$ is drawn when the external resistance is zero: $I_{max} = \frac{E}{r} = \frac{1.9}{380} = 0.005 \; A$.
Since starting a car motor requires a very large current (typically hundreds of amperes),this cell cannot drive the starting motor.