Physics 2002 JAMB Past Questions - Question 32: An electric cell with nominal voltage E has a resistance of 3fi connected across it. If the voltage falls to 0.6E, the internal resistance of the cell is

Correct Answer

C

Explanation

IR=0.61(R+r) ;1*3=0.61(3+r) r=2ohmsWhen an electric cell with a nominal voltage (E) is connected to a resistor (R), the voltage across the resistor (V) can be calculated using Ohm's Law, which is given by the equation:

V = E - I * r

Where:
V = Voltage across the resistor (0.6E in this case)
E = Nominal voltage of the cell
I = Current passing through the circuit
r = Internal resistance of the cell

You mentioned that a resistor with a resistance of 3Ω is connected across the cell. So, we have:

R = 3Ω

Now, we need to find the current (I) passing through the circuit. You can use Ohm's Law for this as well:

I = V / R

Substituting the values we have:

I = (0.6E) / 3Ω

Now, we can substitute this value for I into the first equation:

0.6E = E - I * r

0.6E = E - ((0.6E) / 3Ω) * r

Now, we can solve for r:

0.6E = E - (0.2E) * r

0.6E = E - 0.2E * r

Now, let's isolate 'r' by bringing the terms involving 'r' to one side of the equation:

0.6E - E = -0.2E * r

-0.4E = -0.2E * r

Now, divide both sides by -0.2E to solve for 'r':

r = (0.4E) / (0.2E)

r = 2Ω

So, the internal resistance of the cell is 2Ω.