Question

A cell supplies a current of 0.9A through a 2ohm resistor and current of 0.3A through 7ohm resistor what is the internal resistance of the resistor?

Answer 1

There is a small mistake in the question. There is no term called as internal resistance of the resistor. It must have been 'Internal resistance of the cell'. Anyways, here is how to proceed with this question:

We have the formula for current through a resistor as:
I = $\frac{E}{R+r}$
where 'I' stands for current through the resistor
'E' stands for emf of the cell
'R' stands for value of the resistor and
'r' stands for internal resistance.

On cross multiplication, we get E = I*(R+r)

For the first resistor, we have I = 0.9 A and R = 2 Ω
Hence, E = 0.9*(2+r) 

For the second resistor, I = 0.3 A and R = 7 Ω
Thus, E = 0.3 (7+r)

As the emf of the cell (E) remains constant, we can equate the two equations and hence we get
0.9*(2+r) = 0.3(7+r)
1.8 + 0.9 r = 2.1 + 0.3 r
Shifting variables to one side and constants to another side, we get
0.9 r - 0.3 r = 2.1 - 1.8
0.6 r = 0.3
Multiplying both sides by 10, we get
6 r = 3 or
r = $\frac{3}{6}$ = $\frac{1}{2}$
r = 0.5 Ω

Hence, internal resistance of the cell is 0.5 Ω

Answer 2

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