A dry cell can deliver a current of 0.15A when connected to a resistor. When another resistor is connected in series with the first, the cell can only deliver a current of 0.03A. Calculate the e.m.f and internal resistance of the cell

Answer 1

Let the e.m.f of the cell be E and the internal resistance be r.

When only one resistor is connected, the current delivered by the cell is 0.15A. Using Ohm's Law, we have:

E = (0.15A)(R + r)

Where R is the resistance of the first resistor.

When two resistors are connected in series, the current delivered by the cell is 0.03A. Using Ohm's Law again, we have:

E = (0.03A)(R + 2r)

Now we can set these two equations equal to each other and solve for E:

(0.15A)(R + r) = (0.03A)(R + 2r)

0.15R + 0.15r = 0.03R + 0.06r

0.12R = -0.09r

R = 0.75r

Now we can substitute this in the first equation to solve for E:

E = (0.15A)(0.75r + r)

E = 0.18r

Therefore, the e.m.f of the cell is 0.18V and the internal resistance is 1 Ohm.