Physics 2013 JAMB Past Questions - Question 38: A cell of emf 1.5V is connected in series with a 1 resistor and a current of 0.3A flows through the resistor .Find the internal resistance of the cell

Correct Answer

A

Explanation

To find the internal resistance of the cell, we can use Ohm's Law and the formula for the total resistance in a series circuit.

Ohm's Law states that the voltage across a resistor is equal to the current flowing through it multiplied by its resistance. In this case, the voltage across the resistor is 1.5V and the current flowing through it is 0.3A. Therefore, we can write:

V = I * R

1.5V = 0.3A * 1Ὡ

Now, let's consider the total resistance in the circuit. In a series circuit, the total resistance is the sum of the individual resistances. In this case, we have a 1 Ω resistor and the internal resistance of the cell (let's call it r). Therefore, the total resistance is:

Total resistance = 1 Ω + r

Since the current flowing through the resistor is the same as the current flowing through the cell, we can use Ohm's Law again to relate the total resistance to the current and voltage of the cell:

V = I * (1 Ω + r)

1.5V = 0.3A * (1 Ω + r)

Now, we can solve this equation to find the value of r, the internal resistance of the cell:

1.5V = 0.3A * (1 Ω + r)

1.5V = 0.3A + 0.3A * r

1.5V - 0.3A = 0.3A * r

1.2V = 0.3A * r

r = 1.2V / 0.3A

r = 4 Ω

Therefore, the internal resistance of the cell is 4 Ω