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Soooooo..........
What is it about potential dividers you do not understand that needs clarification?
Q 6(d):
If the resistance wire is linear, then the resistance between both halves at the mid point will be equal. However, theres a matter of that pesky voltmeter internal resistamce.
Can you see how that might alter the voltmeter reading?
in a fixed potential divider you have two fixed resistances (R1 and R2 in series with a supply voltage, Vsupply across the pair.
Because they're in series you know some things. 1. the total resistance Rtotal = R1 + R2 2. the current through both is the same and given by I = Vsupply / Rtotal
the voltage across a resistor is given by ohms law V=IR so the voltage across either resistor is given by Vx=Rx Vsupply/Rtotal --- e.g. if Vsupply=10V R1=7 ohm R2=3 ohm then R total = 7+3 = 10 ohm I = 10/10 = 1 amps V1=R1 * Vsupply/Rtotal =7*10/10 =7
you can repeat for R2 and then confirm that V1+V2=Vsupply as a sanity check
neat.
however it only works if the current in both resistors is the same, so is the current the same when that 5 ohm ammeter is poking about your circuit?
Potentiometers (variable resistor) are nothing more than a variable potential divider.
The two ends of the potentiometer have a resistor connected between them with the resistive element exposed (called the track) so that a third sliding conductor (wiper) is able to move along the track.
The output voltage is picked off by the wiper whose position along the track determines the ratio of the divided resistance.