Question

5. The terminal potential difference of a cell which draws 10A current from a charger is 6.8 V. If the cell
has an open circuit voltage 5.6 V, the internal resistance of the cell is
A) 0.12
B)0.5
C) 0.12 22
D) 0.2 22
E) 0.2422
TL​

Answer 1

Answer:

Using the equation,

V = E+Ir

6.8 = 5.6 + 10r

r = 0.12 ohm

Hence, the answer is option A

Answer 2

The internal resistance of the cell will be 0.12 Ω (A).

Given:

Current in the circuit $=10A$

Terminal potential difference $=6.8V$

Cell potential difference $=5.6V$

To find:

The internal resistance of the cell

Explanation:

• Terminal potential difference is the potential difference present in the circuit when there is current flowing in the circuit.

• The value of the terminal potential difference is less than the $EMF$ of the cell. Terminal potential difference is denoted by $V$.

• The cell potential is the qualitative and quantitative value of potential difference present in the cell even when the current is not flowing through the cell.

• Its value is greater than the potential difference or terminal potential of the circuit and therefore, it has an internal resistance of its own present in the cell. The cell potential or $EMF$ of the cell is denoted by ε.

• The relation between terminal potential difference "$V$", the internal resistance "$r$" of the cell of $EMF$ "ε" is given as

                                      $V=$ ε $-Ir$

           where, $I$ is the current flowing in the circuit.

Solution:

We have,

$V=6.8V$      $;$   ε $=5.6V$       $;$  $I=10A$

Substituting the given values in the equation, we get

$V=$ ε $-Ir$

$6.8=5.6-10r$

$6.8-5.6=10r$

$r=\frac{1.2}{10}$

$r=0.12$ Ω

Final answer:

The internal resistance of the cell will be 0.12 Ω (A).