EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: MarkusAnd on September 08, 2020, 06:32:05 pm
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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
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Hi,
If J3 is a current source, the voltage across J3 is not necessarily = 0.
Jay_Diddy_B
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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?
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Hi,
If I add marker 'D' and I ground A:
[attachimg=1]
(https://www.eevblog.com/forum/beginners/help-with-figuring-out-voltage-in-circuit-analysis/?action=dlattach;attach=1062530;image)
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 = 1A
The current I2 = -1A defined by the voltage at D (-1V) / R2
The voltage at B = R3 x current J3 = 9V
The voltage across J3 D-B = -10V
The voltage between A and C is defined by E4, because E4 is a voltage source.
Is this homework? ;-)
Jay_Diddy_B
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The homework was to solve I1, I2 and I4 8)
I 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 |O
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Hi,
The voltage from D to A = -1V
It is defined by E4 -E5 = 1 -2 = -1
Jay_Diddy_B
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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
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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.
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Hi,
There is an arrow pointing down at the bottom of J3.
[attachimg=1]
So the answer is -10V
Jay_Diddy_B
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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
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There is an arrow pointing down at the bottom of J3.
Yeah - I just noticed that.
I guess that settles it, then. B - A = 9V.
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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
You 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.
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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??
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Hi,
I have used the LTspice circuit simulation tool to model this schematic:
[attachimg=1]
(https://www.eevblog.com/forum/beginners/help-with-figuring-out-voltage-in-circuit-analysis/?action=dlattach;attach=1062606;image)
The node voltages are indicated in purple.
V(B,D) is the difference between V(B) and V(D)
Jay_Diddy_B
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Love it thanks! It's getting easier now :)! So I was thinking right in my last post??