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?
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.
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?
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?
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?
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
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