# Thevenins Theorem

Discussion in 'Homework Help' started by msn56, Apr 18, 2010.

1. ### msn56 Thread Starter Member

Apr 9, 2010
15
0
HI: OK I am missing something here JUST NOT MAKINE SENSE

cALCULATING vTH

SO ACCORDING TO tHEVELINS THEOREM vtH SHOULD EQUAL

vTH = v1*(r1 /(r1 + r2)
oR vTH = 10 (10/15)
= APPROX 6

BUT HERES MY QUESTION. iSNT THE ABOVE A PARALLEL CIRCUIT? aND ISNT VOLTAGE THE SAME IN PARALLEL CIRCUITS? sO HOW CAN THE VOLTAGE BE THE SAME IN PARALLEL CIRCUITS BUT IN THIS PARALLEL CIRCUIT ITS NOT?? WHAT AM i MISSING ?

mIKE

2. ### Ghar Active Member

Mar 8, 2010
655
73
This isn't a parallel circuit.
It's a voltage source in series with a resistor all in parallel with another resistor.
Parallel components exactly share both terminals. The voltage is same across parallel components because you're looking at the same two pieces of wire. Here there's a resistor between the voltage source and your output.

Also, your equation is wrong. The output is across the 5 ohm resistor, making Vth = 10 * 5/15 = 3.33

3. ### msn56 Thread Starter Member

Apr 9, 2010
15
0
Hi Ghar:

Thanks for the answer. Not certain If i am getting it. I reviewed Thevenins theorem here and I still dont see how they calculate the voltage . OK in this example they have

They take out the middle resistor

ok in my mind i redo this as

SO now we have 21volts total resistance is 5 so curretn is 4.2

OK that makes sense so now we have 21 volts leaving and it encounters first resistor

voltage drop would be V = 4.2 (4) = 16.8
yup that makes sense
And if i forget about the T and just treat it like a series circuit then voltage drop at r2 is = 4.2(1) = 4.2 (Or since we started with 21 volts an already lost 16.8 there is only 4.2 left

ok but then we have this "T" and this is where i get confused.

The answer says the voltage is 11.2

I dont see how this is calculated and more importantly trying to logic this through deosn't make sense (gotta b looking at it wrong) .

If I start with 21 volts and I drop 16.8 through the first resistor that only leaves 4.2 after that resistor. Then that voltage gets 'split' at a 'T"

So how could you get 11.2 at one and 4.2 at the other if there is only 4.2 left after the first resistor?

and when we come to the T it looks like a parallel circuit ( I know you said it wasnt ) but is it that the circuit is NOT connected at the T and thats why it is NOT a parralel circuit?

ANother words am I correct in saying this this would be a // circuit

If so then the only difference is that the middle line is not connected and thats why it is not a parallel circuit ?

ANd again how is the 11.2 volts derived there when it looks to me like we only have 4.2 left ?

Sorry for the confusion and i really appreciate your assistance!

4. ### salmanshaheen_88 Active Member

Mar 5, 2009
88
1
why don't ou use mesh and nodal analysis to find the values???

5. ### salmanshaheen_88 Active Member

Mar 5, 2009
88
1
tell me straight away which value you want to calculate???

6. ### msn56 Thread Starter Member

Apr 9, 2010
15
0
Hi Salmanshaheen :
I dont understand how they get Thevelins voltage of 11.2 when they replace the middle resistor - voltage across the middle resistor

Apr 9, 2010
15
0
8. ### Engr Member

Mar 17, 2010
114
5
Try using the Superposition Theorem to solve this circuit. Do not combine the 2 voltage source (B1 and B2) since you are calculating the voltage in between the 2 sources (B1 and B2) to get the Vth.

9. ### Ghar Active Member

Mar 8, 2010
655
73
What you found the voltage drop across the 1 ohm resistor.
However, the voltage you're looking for is the one across the 1 ohm resistor and the 7 V source.

Notice that 4.2 + 7 = 11.2

Alternatively, it's the voltage across the 28 V source and the 4 ohm resistor:
28V - 4.2A*4Ohm = 11.2

Second question:
No, that is not a parallel circuit.
Look at the ends of the resistors.
First resistor is connected across the voltage source. The second resistor is shorted.
The resistors only share 1 terminal. Parallel shares both.