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loading the potential divider?
1) a 100 ohm rotary potentiometer is connected to a 6 v d.c. source with negligible internal resistance. the output required is 3V. The potentiometer is set using a high impedance digital voltmeter connected across the output terminals.
A few minutes later someone else checks the output reading using a moving coil voltmeter which has a resistance of 100 ohms. What is the reading now?
Original voltage divider is
R= R1+R2
where
R = 100 ohms and since voltage is equally divided
R1= R2=50 ohms
By adding a "moving coil voltmeter which has a resistance of 100 ohms" (Never heard of a voltmeter having 100 ohms, but....) we have
Rt= R1+ R2 in parallel with 100 (R3) we have
Rt= R1 + R2R3/(R2 + R3)
Re= R2||R3=R2R3/(R2+R3)=
Re= 50x 100/(50 + 100)
Re = 33 ohms
Rt= R1+ Re= 50 + 33= 83 ohms
Voltage out is product of the circuit current I times the parallel combination of resistors R2 and R3
I= Vin/ Rt
V0= I R3 / (R2+R3)
V0 = (Vin / Rt) Re
V0= (6/ 83) 33= 2.4 V
FlexiPot™: Thin & Flexible Position Sensors