Question

(a) Six lead-acid types of secondary cells each of emf $2.0~V$ and internal resistance 0.015Ω are joined in series to provide a supply to the resistance of 8.5Ω. What is 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?

Answer 1

Hint: Consider e.m.f. As voltage. Use Kirchhoff’s law. Do use faraday's law, since e.m.f terms present in both the formula. Current is nothing but the charge per unit time drawn from voltage or cell. Resistance is known as opposition to the flow of current.
Formula Used:
${{E}_{eq}}=nE$
${{R}_{eq}}=nr+R$
I=E/r

Complete step-by-step solution
In the first question, we have six-cells having a voltage of $2.0~V$ each. The cell's internal resistance is 0.015ohm. Now we need to join this cell in series (not in parallel). Now connect this series with resistance 8.5 ohm. So we need to find out what will be the current flows through resistance due to cell $2.0~V$.
Aim: What is the current drawn from the supply and its terminal voltage?
So our total resistance is given as multiplication of the total cell into 2V.
Mathematically,
${{E}_{eq}}=nE$
${{E}_{eq}}=6\times 2=12V$
Now we have an internal resistance of cell and external resistance. So total resistance is given by
${{R}_{eq}}=nr+R$
Where,
R= internal resistance
R= external resistance
So to calculate current we have,
$I=\dfrac{{{E}_{eq}}}{{{R}_{eq}}}=\dfrac{12}{8.5+6\times 0.015}=1.4A$
Now we need a terminal voltage:
Voltage $=IR=1.4\times 8.5=11.9V$
In the second question, a cell has a voltage of 1.9V and the cell has an internal resistance of 380ohm. Now we need to calculate the current drawn from the cell and find out that the given cell can help to drive the starting motor of a car.
Aim: What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
$E=1.9V,\text{ }r=380ohm$
$I=\dfrac{E}{r}=\dfrac{1.9}{380}=0.005A$
To start the motor of a car at least 100A is required for a few seconds. Calculating the value of current is very much less so it won’t help the car to drive.

Note: Terminal voltage is nothing but voltage between two points or voltage from one point of the cell to another point of the cell. In the first case, do not consider internal resistance in terminal voltage because internal resistance is concerned with cells only. Because the cell is a device and each device has its own internal resistance.