Help with figuring out voltage in circuit analysis. - Page 1
Forum section: electronics / beginners

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 06:32:05 pm
content: Hi!Just started a basic course in circuit analysis and came across this circuit(it's attached to this post).My problem with the circuit is this:The current(J3) going through R3 is 1A. Because of ohm's law the voltage drop over R3 is 9V. So the potential difference between point A and B is 9V.Therefore because the voltage over E5 is 2V the potential difference between point A and C should be 11V, but E4 is only 1V??Can someone explain why the potential difference between A and C = E4 and not 11V??// Markus

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 08, 2020, 06:38:52 pm
content: Hi,If J3 is a current source, the voltage across J3 is not necessarily = 0.Jay_Diddy_B

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 06:46:50 pm
content: Okay but the voltage on the node between J3 and E5 is -1V, so if the voltage over R3 is 9V the voltage over J3 should be -8V right?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 08, 2020, 06:56:03 pm
content: Hi,If I add marker 'D' and I ground A:The voltage on C is 1V (defined by E4)The voltage on D is 1V - 2V = -1V (defined by E4 and E5)The current I1 = 1A defined by E4/R1 = 1AThe current I2 = -1A defined by the voltage at D (-1V) / R2The voltage at B = R3 x current J3 = 9VThe voltage across J3 D-B = -10VThe voltage between A and C is defined by E4, because E4 is a voltage source.Is this homework? ;-)Jay_Diddy_B

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 07:10:43 pm
content: The homework was to solve I1, I2 and I4I still don't get why J3 is -10V and not -8V? If the voltage drop over R3 is 9V and D = -1V:The potential from D to A is -1V-8V+9V = 0V

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 08, 2020, 07:17:12 pm
content: Hi,The voltage from D to A = -1VIt is defined by E4 -E5 = 1 -2 = -1Jay_Diddy_B

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 07:22:50 pm
content: Okay so what i meant was that:A is grounded(0V), D is -1V, if I want to go from -1V to 0V with the R3 voltage(9V) taken into account the voltage over J3 should be -8V because -1V-8V+9V = 0V

author: ledtester
author_rank: Super Contributor
timestamp: September 08, 2020, 07:35:25 pm
content: The direction of the current source J3 is not specified. Usually there is an arrow on current sources so you know which direction the current is flowing - just like there are + and - signs on voltage sources.As it stands, the voltage difference B-A could be 9V or it could be -9V depending on if the 1A is flowing "up" or "down" in the schematic.If B-A is 9V, then the voltage drop across J3 (from top to bottom) is -1 - 9 = -10V.If B-A is -9V then the voltage drop across J3 is -1 - -9 = 8V.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 08, 2020, 07:39:47 pm
content: Hi,There is an arrow pointing down at the bottom of J3.So the answer is -10VJay_Diddy_B

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 07:41:38 pm
content: Ok it feels like I'm arguing now but i just cant get my head around it how the voltage at B = -10V if the voltage over R3 is 9V.Shouldn't the voltage at B be -9V?? Because if the voltage at B is -10V and the voltage over R3 is 9V the voltage at A should be -10V+9V = -1V

author: ledtester
author_rank: Super Contributor
timestamp: September 08, 2020, 07:42:25 pm
content: QuoteYeah - I just noticed that.I guess that settles it, then. B - A = 9V.

author: ledtester
author_rank: Super Contributor
timestamp: September 08, 2020, 07:46:49 pm
content: Quote from: MarkusAnd on September 08, 2020, 07:41:38 pmYou always have to specify where your 0V potential is.If A = 0V, then B = 9V and the voltage at the negative side of E5 is -1V. The voltage at C is 1V.(Note how I am making A my 0V potential point.)The voltage across J3 is -10V.

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 07:57:28 pm
content: Waait I think i got it now!So the voltage goes like this from C to A:C to D: 1V - 2v = -1V(at D)D to B: -1V - (-10V) = 9V(at B)B to A 9V - 9V = 0V(at A)Right??

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 08, 2020, 08:00:03 pm
content: Hi,I have used the LTspice circuit simulation tool to model this schematic:The node voltages are indicated in purple.V(B,D) is the difference between V(B) and V(D)Jay_Diddy_B

author: MarkusAnd
author_rank: Contributor
timestamp: September 08, 2020, 08:10:30 pm
content: Love it thanks! It's getting easier now! So I was thinking right in my last post??