The cell in the circuit has an emf of 2.0 V. When the variable resistor has a resistance of 4.0 Ω, the potential difference (pd) across the terminals of the cell is 1.0 V.
What is the pd across the terminals of the cell when the resistance of the variable
resistor is 12 Ω?
The answer is 1.5 V
I've got no Idea how they got there, please would someone help?
Really stuck on this electricity physics question. Please help! watch
- Thread Starter
Last edited by BigChicken; 15-03-2016 at 23:02.
- 15-03-2016 22:46
- Official Rep
- 18-03-2016 00:03
Sorry you've not had any responses about this.
Why not try posting in a specific subject forum- you might have more luck there.
Here's a link to our subject forum which should help get you more responses.
- 18-03-2016 11:26
First we calculate the current of the circuit and resistance of the internal resistor:
V/R = I = 1/4 = 0.25A
Internal resistance = 1/0.25 = 4ohms
Next we work out the output voltage with 12ohm variable resistor:
Vout = Vint x R1/R1+R2 = 2 x 12/12+4 = 1.5V
There you go.