Circuit calculation - Page 1
Forum section: electronics / beginners

author: Irukandji
author_rank: Contributor
timestamp: June 22, 2021, 10:18:37 pm
content: Hi guys,On the picture you see a circuit. The Voltage between the four 330 Ohm resistors must be 0 V. I draw it in this picture. When there is no Voltage i have no ampere between the 100K Ohm potentiometer and the first Dot between the two 330 Ohm resistor or ? I mean the ampere measurement with the red "" ?1 "" Symbol.Is my calculation right?U1 = 330 Ohm * 0,02A = 6,6VU2 = 330 Ohm * 0,02A = 6,6VU3 = 330 Ohm * 0,02A = 6,6VU4 = 330 Ohm * 0,02A = 6,6VU1,2= 6,6V+6,6V ? 13,2VU3,4= 6,6V+6,6V ? 13,2VU = 13,2VMy calculation for the second Question Symbol"" ?2 ""I=U/R = 13,2V / 100000Ohm = 1,32*10^-4 A= 0,132 mA( I ) for the circuit is then:I = U/R = 13,2V / 328,9 Ohm = 0,04 AIs my calculation right, or did I make a big mistake?

author: IamSynthetiC
author_rank: Regular Contributor
timestamp: June 23, 2021, 12:51:38 am
content: No,Voltage is a form of potential (or dynamic) energy.Newton's seconds lawstates that dynamic energy is anindefiniteintegral, whitch means that V = (solution to integral) + C (C is an arbitary constant or Constant of Integration i think it is called in english). All that means that one can NEVER talk about potential energy (or Voltage) without defining aZero Level(or ground). Also, since one can choose any Zero Level, voltages like 1,2,3 or 0,-1,-2,-3 are all perfectly valid.Now to your circuit.If the voltage between the resistors is 0V (when voltage is been applied), then that means that your ground is not really 0V and that your supply is supplying 6.6V on the plus side (as you correctly found), but -6.6V in the minus side. (since the same current flows through the resistors, there is going to be a voltage drop (delta V) dV = I * R = 6.6V. So there is a 6.6V drop from the plus side to the center (which is at 0V, and another 6.6V from the center (which is at 0V) to the minus side, therefore yourGNDis at -6.6V).The current through the potentiometer you have found is correct (I = dV/R => I = (6.6V - (-6.6V))/100kR => I = 13.2 * 10^-5)But now you see, while the voltage between the resistors is 0V, the voltage on the low side is -6.6V, (dV = (0 - (-6.6V)) = 6.6V) therefore there is current flow through the lower side of the potentiometerEither that is the case, ori have seriously misunderstood the problem.Hope that helps and good luck.

author: Ian.M
author_rank: Super Contributor
timestamp: June 23, 2021, 01:27:48 am
content: @IamSynthetiC: WTF?@all,The circuit shown has no connection to ground. Its floating with no pre-determined reference potential. Its therefore up to us to chose an arbitrary reference node in the circuit, that is most convenient for solving the problem. In the absence of any other considerations, with a single DC voltage source, its common to use its negative side as the 0V reference point. Let us do so.The known current through U1,U2 lets us calculate the source voltage (13.2V), and the voltage at their junction node (6,6V). Knowing the voltmeter reads 0V, and that U3,U4 are identical to U1,U2, means that the current through ammeter #1 must be zero, as any non-zero current would shift the potential at the U3,U4 junction node resulting in a non-zero voltmeter reading.Therefore the potentiometer U5 must be set at exactly 50% so the upper and lower sections of its potential divider are equal, for a Thévenin-equivalent voltage VThof 6.6V, and with no wiper current, the current through its track is solely determined by the voltage across it and its track resistance giving A2 of 13.2V/100K = 132uA.Determining the total current from the voltage source is then trivial as all branch currents are now known.

author: Nusa
author_rank: Super Contributor
timestamp: June 23, 2021, 04:51:11 am
content: Quote from: Irukandji on June 22, 2021, 10:18:37 pmYou did it exactly right, in theory. That final calculation could be replaced with the trivial sum of all branches to get 40,132 ma. That's almost the same number you get from your resistance equivalent calculation(I think you rounded there too) if you don't round away the significant part of the answer.The only mistake, so far as a homework problem is concerned, is how you rounded numbers when doing math. Why should one current calculation have a result in micro-amp precision, while others round to milli-amp precision? The fact you expressed units in amps for Ohms law purposes doesn't change the need for a consistent level of accuracy to demonstrate that you've solved the entire circuit.

author: Irukandji
author_rank: Contributor
timestamp: June 23, 2021, 02:26:44 pm
content: Hi,thanks for your answers.So sorry to forget the ground symbol. I have drawn the ground symbol now on the bottom left side. I uploaded now two pictures, the first picture is the old picture with a ground symbol.In the second picture you see in line with the potentiometer a next 330 Ohm resistor and a current from 20 mA.But this can never happen, right? If I have 0 V again between the four 330 Ohm resistors there can be no current on the resistor between potentiometer and the ampere meter, right? So it must be a trap, right?

author: IamSynthetiC
author_rank: Regular Contributor
timestamp: June 23, 2021, 03:27:07 pm
content: oupsI am terribly sorry, in my spleepyness i completly misread the cirquit. Disregard everything i said about calculations.In your second picture, current flowing through the resistor (Lets call it U6) means there is a voltage difference at its terminals (Vd = I*R = 0.02*330 = 6.6V). That means that the middle of the potentiometer has 13.2V, which in turn means the potentiometer has no top resistance

author: Terry Bites
author_rank: Super Contributor
timestamp: June 23, 2021, 03:36:52 pm
content: Yes you did it right! You dont need a ground symbol as long as you decide on a fixed referrence point. It doesnt have to be static either, you could put 100Vdc or 240Vac (don't) on the negative side of you battery and still get the same answer. You only need ohms law. Invoking Newton and or any calculus is pure bs.

author: Ian.M
author_rank: Super Contributor
timestamp: June 23, 2021, 07:07:50 pm
content: Your 'circuit_new' from reply #4 is impossible. U1, U2 and the current through them remain the same, giving 13.2V for the voltage source, however to maximize*the current through A1, the potentiometer wiper must be set at the top of its track so the new resistor is effectively in parallel with U3. As two equal resistors in parallel have a total resistance of half of the individual resistances, there can only be one third of the supply voltage across U3 and across the new resistor. However we know there must be 6.6V across it to get 20mA through it, which would require a 19.8V voltage source, which is inconsistent with the 13.2V required for 20mA through U1,U2.* If the wiper was not at the top of the potentiometer track an even higher voltage would be required to get 20mA through A1 in the direction shown.P.S. On schematics it is conventional to reserve designators starting with U for integrated circuits or hybrid modules, and to use R for resistors. Seehttps://en.wikipedia.org/wiki/Reference_designator