Question

Kindly help me? thanks a bunch! :)

Answer 1

When a particular cell is on open circuit, the p.d. between its terminals is 1.5 V. When a 10 Ω resistance is connected between the terminals the p.d. falls to 1.0 V and when the 10 Ω resistance is replaced by a resistance R the p.d. becomes 0.5 V. What is: a) the internal resistance of the cell; b) the value of R?
thank you! i really am confused at the moment !

Answer 2

The 1.5V would be the emf of the cell (no load/resistance in the circuit). Once the cell is in a circuit, with 10Ω resistance, you notice the p.d fall to 1.0V. The 0.5V missing is being used by the internal resistance of the cell.
Let V equal the voltage when there is a resistance in there (the 10Ω here), and v = the lost p.d due to internal resistance.
Use the equation
emf = v + V = Ir + IR = I (r + R)

first you need the current - only achieved when the cell is in a loaded circuit. I = V/R so I = 1.0 / 10 = 0.1A
You should now be able to find the value for r (the internal resistance).

For part b, r is still the same (its inherent to that cell), and you know it is now using up 1V. (i.e V = 0.5V, using emf = v + V = 1.5 = v + 0.5) . Use this v to find the new current when R is in the circuit, as before (I = v / r)

now, the emf is the same and you have a new I. So:
emf = 1.5 = I (5+R)
solve for R.

Hope that helps

Answer 3

It does help! thanks alot! but there's one problem, im sill clueless as to how to calculate R. I'm not a pro at this...

Answer 4

did you calculate a new value for I in part b) ? you should get 0.2A. Then it should become easy..?

Answer 5

i don't understand anything :(