A cell supplies current of 0.6A and 0.2A through 1ohms and 4.0ohms resistor respectively. Calculate the internal resistance of the cell

0.5Ω

Since different currents are passing through the resistors, then the resistors are most probably connected in parallel. This also means that the same voltage will pass across them.

Using Ohm's law, the voltage across a resistor in a circuit is given by;

V = I(R + r)            -----------(i)

For the 1ohm resistor, the voltage across it is given by;

Where;

I = current passing through the 1 ohm resistor = 0.6A

R = resistance of the 1 ohm resistor = 1Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.6(1 + r)                 -------------------(ii)

For the 4.0ohm resistor, the voltage across it is given by;

Where;

I = current passing through the 4.0 ohms resistor = 0.2A

R = resistance of the 4.0 ohms resistor = 4.0Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.2(4.0 + r)                 -------------------(iii)

Now solve equations (ii) and (iii) simultaneously;

V = 0.6(1 + r)

V = 0.2(4.0 + r)

Substitute the value of V in equation (ii) into equation (iii). Therefore, we have;

0.6(1 + r) = 0.2(4.0 + r)

Solve for r

0.6 + 0.6r = 0.8 + 0.2r

0.6r - 0.2r = 0.8 - 0.6

0.4r = 0.2

r =

r = 0.5

Therefore, the internal resistance of the cell is 0.5Ω

1/0.14nm = 10^10/1.4 a lot

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