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Really stuck on this electricity physics question. Please help!

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?
[1 mark]

The answer is 1.5 V

I've got no Idea how they got there, please would someone help?
(edited 8 years ago)
Sorry you've not had any responses about this. :frown:

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. :redface:

:h:
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.
Reply 3
You don't even need to work out the current, i f the v across the cell is is same as the v across the resistor then they will have the same resistance because they're in series. So r is 4 ohms on both.
Original post by dylan1016
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.

Why is it 2 x 12 and not 2 x 4 as isn't 4 the resistance of the battery not 12 and the question asks for the pd across the battery terminals?
I know this sounds stupid but I can’t wrap my head around why Vin is the emf when the voltage provided is actually less than the emf. Anyone help me out?
Original post by Elliehobson0805
I know this sounds stupid but I can’t wrap my head around why Vin is the emf when the voltage provided is actually less than the emf. Anyone help me out?

Never mind guys I figured it out!
I know this sounds stupid but I can’t wrap my head around why Vin is the emf when the voltage provided is actually less than the emf. Anyone help me out?
does potential difference across the terminals just refer to the terminal pd?
Reply 9
Original post by Elliehobson0805
I know this sounds stupid but I can’t wrap my head around why Vin is the emf when the voltage provided is actually less than the emf. Anyone help me out?
because the terminal pd (1.5) is the pd given out after energy loss in the battery which is lost volts
Reply 10
Original post by Elliehobson0805
I know this sounds stupid but I can’t wrap my head around why Vin is the emf when the voltage provided is actually less than the emf. Anyone help me out?
Because the Terminal pd (1.5V) is the pd on the circuit outside the battery so it is the pd given to the rest of the circuit after the EMF (Vin) experiences energy loss thats why you call it lost pd

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