then shouldn't the variable resistor be connected to the negative terminal of the battery?
It makes no difference at all where the variable resistor is connected.
The current will be given by Ohms law. viz
i = V/R.
Regardless of where the resistor is located in the circuit shown the electrons pass through it in the same direction, and the same voltage drop occurs in it.
There are circuits where the order of the components matters but that is due either to a branch in the path making conditions different at different locations in the circuit, or due to special interactions outside the scope of Ohm's law (eg magnetic fields may cause interactions between components. )
I'm a bit new to electricity so could you explain intuitively why it doesn't matter where it's located? (Where does my reasoning fail?) Thanks! –
Consider this water analogy.
Remember "ALL models are wrong. Some models are useful" - George Box.
A pump with a known pressure/flow rate output characteristic is run at a fixed power input level. A standard table says that it operated with open output into a tank with zero head it will pump 1000 gallons per minutes.
The same table says that if it pumps into a 4" pipe 80 feet long, followed by a 2" pipe 20 feet long followed by a 1" pipe 5 feet long then exits with a 3 foot head above the inlet level it will have say a 100 gallons per minute flow rate.
Note, I made the pipe lengths and diameters up - but each mayt have ABOUT the same hydrilic drop - maybe not, not important.
80' x 4" + 20' x 2" + 5' x 1" + 3 foot head = 100 gpm
What flow rate would you expect if the pipes were kept the same but reordered eg
3' head exit in all cases plus
80 20 5 = 100 gpm
80 5 20 = ?
20 5 80 = ?
20 80 5 = ?
5 80 20 = ?
5 20 80 = ?
I strongly suspect that you'll find that the results will be very close to the same in all cases. In the resistive circuit case the results are exactly the same for ideal resistors and real resistors are close to ideal. In pipes with water here may be some second order effects at boundaries but the example is close enough.
Using Ohms's law:
Voltage drop = current x resistance it flows in.
V = I x R
I = V/R
R = V/I
In this case total resistance = R_Resistor + Rwire.
R wire is so low (usually) that it can be ignored BUT even if it cannot be the result is still the same whatever the order.
R_Resistor only - ignore R_wire.
I = V/T = V/ T_Resistor - current will be such that the drop around the circuit balances Vbattery. If the resistive drop is lower than Vbattery then current will increase until it balances. The battery does not "know or care" where the resistor is in the circuit - it outputs (conventional current from V+ and accepts the return at V-.
R_Resistor (Rr) +R_wire Rw
2a Flow is through Rr and then through Rw.
Vrr = IR = I x Rr
Vrw = IR = I x Rw
Vtotal = Vbat = I x Rr + I x Rw = I x (Rr + Rw) ....(1)
2b Flow is through Rw and then through Rr.
Vrw = IR = I x Rw
Vrr= IR = I x Rr
Vtotal = Vbat = I x Rw + I x Rwr= I x (Rw + Rr) ....(2)
BUT Rtotal in circuit = Rr + Rw = Rw + Rr = Rtotal
Ra = RA and
Rw = RW and
Vbattery1 = Vbattery2
then the current is IDENTICAL in both cases. then
Equation (1) and (2) are identical.
The total drop is the result of the common current through Rr + Rw or Rw + Rr. The battery does not 'know' or 'care' which comes first.
Current which flows depends only on the total opposition to current flow.