Q. 3.14. (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?

Question

Accepted Answer

(a) Number of secondary cells, $n=6$

Emf of each secondary cell, $\mathrm{E}=2.0 \mathrm{~V}$ Internal
resistance of each cell, $\mathrm{r}=0.015 \Omega$ series resistor is
connected to the combination of cells.
Resistance of the resistor, R = 8.5
Current drawn from the supply $=\mathrm{I}$, which is given by the relation,
$$
\begin{aligned}
I & =\frac{n E}{R+n r} \
& =\frac{6 	imes 2}{8.5+6 	imes 0.015} \
& =\frac{12}{8.59}=1.39 \mathrm{~A}
\end{aligned}
$$

Terminal voltage, $\mathrm{V}=\mathrm{IR}=1.39 	imes 8.5=11.87 \mathrm{~A}$
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A .
(b) After a long use, emf of the secondary cell, $\mathrm{E}=1.9 \mathrm{~V}$

Internal resistance of the cell, $\mathrm{r}=380 \Omega$
Hence, maximum current $=\frac{E}{r}=\frac{1.9}{380}=0.005 \mathrm{~A}$
Therefore, the maximum current drawn from the cell is 0.005 A . Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.