This equation summarises Ohm’s law. It suggests that any value of voltage you put across a resistor divided by the current it produces in the resistor, will always give the same value of resistance. So, if you plotted a graph you would get:
Current used in all 3 will be the same (current doesn’t get used up) but energy used per coulomb (i.e. pd) will depend on value of resistance
So,
Here the voltage across all three will be the same but current through each depends on resistance of each.
So:-
Therefore, inside the cell you get some energy put IN to the circuit by the cell (an emf) some energy taken OUT of the circuit by the resistor (a pd).
So the pd, V, available to the rest of the circuit (the external circuit, as some questions ...
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Current used in all 3 will be the same (current doesn’t get used up) but energy used per coulomb (i.e. pd) will depend on value of resistance
So,
Here the voltage across all three will be the same but current through each depends on resistance of each.
So:-
Therefore, inside the cell you get some energy put IN to the circuit by the cell (an emf) some energy taken OUT of the circuit by the resistor (a pd).
So the pd, V, available to the rest of the circuit (the external circuit, as some questions may refer to it) is:
V = E - Ir
where:
E = the emf of the cell
I = the current through the cell
r = the value of the internal resistance
so
Ir = the p.d. across the internal resistor.
Note - V is sometimes called the terminal pd as it is the pd across the terminals - clever, hey!
Example
What is the terminal p.d. for a cell of emf 2V and internal resistance 1 ohm when it is connected to a 9 ohm resistor?
Answer/p>
To work out impact of this, just pretend the internal resistance is just one of the normal resistors in the circuit. Draw it in the circuit diagram next to the cell so that all the current that goes through the cell also goes through the resistor.
So to find V, the terminal pd (or the voltage available to the external circuit), calculate current I for the whole circuit:
Note - VT and RT are the voltage and resistance for the whole circuit, including external and internal resistance.)
Therefore, the 9 resistor gets V = IR = 0.2 x 9 = 1.8V
So this 2V emf cell actually supplies 1.8V to the external circuit.
Now, swap the 9 resistor for a 1.
Find V, the terminal pd, using the same method again:
The perfect ammeter must have zero resistance.
The perfect voltmeter must have infinite resistance.
Otherwise, when you put them into the circuit they would change the circuit.
In reality they aren’t perfect. So in the circuit above, some of the emf supplied to the circuit is used pushing current through the ammeter, and some of the current goes through the voltmeter, not through R.
How can you deal with this? Well, pretend that the voltmeter is actually not a voltmeter but just a resistor and look at the effect it has.
Example:
What should the current, I, be in this circuit if the voltmeter is
(a) perfect (infinite resistance)
(b) 20 k resistance?
