Question

Six lead-acid types of secondary cells each of emf 2.0V and internal resistance $0.015\,\Omega$ are joined to provide a supply to a resistance of $8.5\,\Omega$ . What is the current drawn from the supply and its terminal voltage?
(A) $1.4\,{\text{A}}$ , $11.9\,{\text{V}}$
(B) $1.8\,{\text{A}}$ , $11.9\,{\text{V}}$
(C) $1.4\,{\text{A}}$ , $12.9\,{\text{V}}$
(D) $1.8\,{\text{A}}$ , $12.9\,{\text{V}}$

Answer 1

First of all, we will apply a formula to calculate current which is net emf over total resistance present in the circuit. again, to calculate the terminal voltage we will use the Ohm’s law.

Complete step by step solution:
In the given problem, we are supplied with following data:
There are six lead-acid types of secondary cells.
The emf of each cell is $2.0\,{\text{V}}$ while the resistance of each cell is $0.015\,\Omega$ .
The external resistor has a resistance of $8.5\,\Omega$ .
We are asked to find the current drawn from the supply and its terminal voltage.
To begin with, let us analyse the setup, in which we are given that there are six cells which are connected in series. So, the resistance of all the cells combined will increase to a higher value. Same is the case with the emf too, as they are connected in series, the total emf of all the cells combined will increase.
Now, we proceed to solve the problem:
We know, a formula which gives the total amount of current in the circuit with the cells connected, is given by:
$I = \dfrac{{nE}}{{R + nr}}$ …… (1)
Where,
$I$ indicates the current in the circuit.
$n$ indicates the number of cells.
$E$ indicates the emf of each cell.
$R$ indicates the values of the external resistance.
$r$ indicates the resistance of each cell.
Now, we substitute the required values in the equation (1) and we get:
I = \dfrac{{nE}}{{R + nr}} \\\ \Rightarrow I = \dfrac{{6 \times 2.0}}{{8.5 + 6 \times 0.015}} \\\ \Rightarrow I = 1.4\,{\text{A}} \\\
Therefore, the current drawn from the supply is $1.4\,{\text{A}}$ .
Again, we can find the potential difference by simply using the Ohm’s law:
$V = IR$ …… (2)
Where,
$V$ indicates the potential difference.
$I$ indicates the current drawn from the supply.
$R$ indicates the external resistance of the circuit.
Now, we substitute the required values in the equation (2) and we get:
$V = IR \\\ \Rightarrow V = 1.4 \times 8.5 \\\ \Rightarrow V = 11.9\,{\text{V}}$
Therefore, the terminal voltage is found to be $11.9\,{\text{V}}$ .

The correct option is (A).

Note: While solving the problem, it is important to remember that the potential difference is used to measure the energy between two certain points in the circuit whereas the emf is characterised as the maximum voltage which a battery delivers. Magnitude of potential difference is always less than that of emf of a battery.