- 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?
- 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?
Exercise
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- Number of secondary cells, n = 6
Internal resistance of each cell, $\text{r}=0.015\ \Omega$
series resistor is connected to the combination of cells.
Resistance of the resistor, $\text{R}=8.5\ \Omega$
Current drawn from the supply = I, which is given by the relation,
$\text{I}=\frac{\text{nE}}{\text{R}+\text{nr}}$
$=\frac{6\times2}{8.5+6\times0.015}$
$=\frac{12}{8.59}=1.39\ \text{A}$
Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
- After a long use, emf of the secondary cell, E = 1.9 V
Hence, maximum current $=\frac{\text{E}}{\text{r}}=\frac{1.9}{380}=0.005\ \text{A}$
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.
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