Physics 2001 JAMB Past Questions - Question 36: A cell can supply currents of 0.4 A and 0.2 A through a 4.CQ and 10.0Q resistors respectively. The interna! resistance of the cell is

Correct Answer

E

Explanation

E=0.4(4+r)and E=0.2(10+r)r=2ohmsTo find the internal resistance of the cell, you can use Ohm's law, which states that the voltage (V) across a resistor is equal to the current (I) passing through it multiplied by its resistance (R):

V = I * R

In your case, you have two resistors with currents of 0.4 A and 0.2 A. Let's call the internal resistance of the cell "r" ohms, and the resistors "R1" and "R2":

For the first resistor:
V1 = 0.4 A * (4 Ω + r)

For the second resistor:
V2 = 0.2 A * (10 Ω + r)

Now, you know that the total voltage supplied by the cell is the same for both resistors:

V1 = V2

So, you can equate the expressions for V1 and V2:

0.4 A * (4 Ω + r) = 0.2 A * (10 Ω + r)

Now, you can solve for the internal resistance (r):

0.4(4 + r) = 0.2(10 + r)

1.6 + 0.4r = 2 + 0.2r

0.4r - 0.2r = 2 - 1.6

0.2r = 0.4

r = 0.4 Ω

So, the internal resistance of the cell is 0.4 ohms.