help with this simple nodal analysis issue, please

[PG1995]

Hi,

1: https://img19.imageshack.us/img19/2256/nodaldiagram1.jpg (diagram of circuit and few other details)
2: https://img36.imageshack.us/img36/5405/nodaldiagram2.jpg (solution)

The book asks "Find the magnitude and polarity of the voltage across each resistor".

I was trying to find the magnitude and polarity across R3 in the circuit diagram (please have a look on the link #1).

The values for V1 and V2 are correct. V1 is -14.9v, and V2 is -12.7v. The magnitude for voltage across R3 I found was 14.8v, but the book says it's 9.71v.

I4 is -3.7A (the assumed direction by me was from left to right of screen as is shown in the scan).

The book also says: V_R3 = V1 + 12 -V2. I don't get this step. Is the book applying KVL here, then how. Please help me with this. Many thanks for the help.

Cheers

 

[Ratchit]

PG1995,

The values for V1 and V2 are correct. V1 is -14.9v, and V2 is -12.7v. The magnitude for voltage across R3 I found was 14.8v, but the book says it's 9.71v.

The book is correct. See the calculation for I4 in the attached file.

The book also says: V_R3 = V1 + 12 -V2. I don't get this step. Is the book applying KVL here, then how. Please help me with this.

The book is correct. See the VR3 calculation in the attachment.

Ratch

 

[dougy83]

Ok, so you've got the voltages at points V1 and V2 of -14.86 & -12.57. Now work out the voltage at the node connecting R3 & E, call it Vx. The voltage across R3 is Vx-V1, which is V2-E-V1. The second form I've just written is just the negative of what the book says; i.e. V2-E-V1 = -(-V2+E+V1) = -(V1-12-V2). Because you've been asked for the 'magnitude' of the voltage, you just flip the polarity to make it positive.

To find Vx; Vx = V2 - 12 = -24.57
The voltage across R3 is: Vx-V1 = -24.57 - -14.86 = -9.71, the magnitude is 9.71

 

[PG1995]

Ratch, thanks a lot.

From your PDF: Voltage across R3, assuming V1 more positive, 9.714

Now I see my value of current for I4 is wrong. Okay. But please have a look on link #2 in the first post of mine above (please have a look at the bottom the part highlighted orange). V1 is actually more negative than V2. V1 is -14.9v, and V2 is -12.7v. Then, how can you assume V1 is more +ve than V2? And, is the book applying KVL in some way to find the voltage across R3? But how? Please guide me. Thanks a lot.

 

[PG1995]

The second form I've just written is just the negative of what the book says; i.e. V2-E-V1 = -(-V2+E+V1) = -(V1-12-V2). Because you've been asked for the 'magnitude' of the voltage, you just flip the polarity to make it positive.

To find Vx; Vx = V2 - 12 = -24.57
The voltage across R3 is: Vx-V1 = -24.57 - -14.86 = -9.71, the magnitude is 9.71

Thanks a lot, dougy. I think I'm close to the answer. I just saw your reply now.

But I don't understand why you had to flip the polarity to make it positive. When you have -(-V2+E+V1) then what's the need to make it V_R3=V1+12-V2 (the book form)?

 

[dougy83]

Because when asking for the magnitude of a voltage you are being asked to neglect the polarity. If I say the voltage across R3 is Vx-V1, my result is negative, if I say V1-Vx, my result is positive.

Sorry, I didn't read your whole post; it asks for both the magnitude and polarity of the voltage across each resistor. Well, you know the magnitude already - it's the positive 9.71. As for the polarity, V1 is -14.9V & Vx is -24.6V, so the left terminal of R3 is more positive.

 

[PG1995]

Many, many thanks, dougy.

Now I'm a bit confused. Okay. But please have a look on link #2 in the first post of mine above (please have a look at the bottom the part highlighted orange). The current flows from left of screen towards right (towards V2) - at least this was my original assumed current. The current flows from higher potential to lower. So, don't you think Vx, the point between R3 and 12V supply, should be subtracted from V1 because current flows from it which means higher potential. i.e. V1 - Vx. Then, the current flows through the 12V supply toward V2 which means V2 is lower potential point, then won't we right 12-V2. Then, Vx=12-V2. The end form would be: V1-Vx = V1-(12V-V2) = V1-12V+V2. Please guide me on this.

 

[Ratchit]

PG1995,

But please have a look on link #2 in the first post of mine above (please have a look at the bottom the part highlighted orange). V1 is actually more negative than V2. V1 is -14.9v, and V2 is -12.7v. Then, how can you assume V1 is more +ve than V2? And, is the book applying KVL in some way to find the voltage across R3? But how? Please guide me.

Good question. I am calculating the voltage referenced from the junction of R3 and the 12 volt source. Therefore that voltage is subtracted from the voltage on the other end of R3 at node 1. So, V1 = -14.86 from which we subtract (V2-12) = -24.57 . Subtracting we get -14.86 - (-24.57) = 9.71 . Therefore the voltage across R3 is 9.71 volts positive referenced from the junction of R3 and the 12 volt source. Notice that this is not the same as the voltage between node 1 and node 2.

Ratch

 

[dougy83]

Now I'm a bit confused. Okay. But please have a look on link #2 in the first post of mine above (please have a look at the bottom the part highlighted orange). The current flows from left of screen towards right (towards V2) - at least this was my original assumed current. The current flows from higher potential to lower. So, don't you think Vx, the point between R3 and 12V supply, should be subtracted from V1 because current flows from it which means higher potential. i.e. V1 - Vx. Then, the current flows through the 12V supply toward V2 which means V2 is lower potential point, then won't we right 12-V2. Then, Vx=12-V2. The end form would be: V1-Vx = V1-(12V-V2) = V1-12V+V2. Please guide me on this.

Sorry for causing the confusion. If you assume the current flows from left to right (as you have), you get a positive value for the voltage across the resistor.

I've attached the highlighted circuit from the 2nd page you mentioned, but shown in a loop so that KVL can be easily applied.

If voltage loop solved anticlockwise (from 0V), current assumed to be anticlockwise - which is the same as from left to right in your original drawing:
14.9+VR-12-12.7 = 0
VR = 12 + 12.7 - 14.9 = 9.8

If voltage loop solved clockwise (from 0V), current assumed to be clockwise:
12.7+12+VR-14.9 = 0
VR = 14.9 - 12 - 12.7 = -9.8

So you get a different sign depending on which way you assume the current to flow; but in each case, the left terminal of the resistor is 9.8V more positive than its right terminal.

I hope this makes sense and doesn't add to your confusion..

 

[PG1995]

Many, many thanks, both of you. I offer my deepest gratitude to both of you. dougy, your attached circuit diagram made things clear. Thanks. If I have any follow-on questions, will ask you again.

With best wishes
PG