Thevenin's Voltage in Circuit

[ampegbassplayer1969]

Hey guys -

I'm taking night classes to learn electronics and I'm stuck on this question...

I need to calculate Thevenin's voltage across points A and B (The node where the 510 Ohm load is connected in the picture), with the 510 Ohm load disconnected. The answer is supposed to be 2 volts, but I can't figure it out... I calculate Thevenin's resistance as 1137 Ohms which I'm confident is correct. Could anyone walk me through calculating VTh?

Thanks for your help.

I apologize if this is in the wrong section!

 

[audiomixer]

to find the answer you might try to find the voltage across [R1 + R2].... when you have that Ohm is your friend and you get the voltage drop across these.

you might also redraw the circuit so that the layout:



- michael

 

[JohnRoberts]

I don't know about Thevenin... but with 510 ohm disconnected, there is no current flowing through r3 (470 ohm) so no voltage drop across it.  Voltage at junction of R1 and R2 is simple divider between 1.5V and 3V voltage sources.

More complicated with 510 Ohm connected.

JR

 

[ampegbassplayer1969]

Using Superposition Theorem:

VS1 becomes R1 + R2 in series with 1.5v source. The 3 volt source is replaced with its internal resistance of 0 ohms.

VT = 3 Volts
RT = 3k Ohms
IT = 1.5/3k = 500uA
I1 = 500uA
I2 = 500uA
V1 = 500mV
V2 = 1V

NOTE: The current flowers counter-clockwise

VS2 becomes R1 + R2 in series with 3v source. The 1.5 volt source is replaced with its internal resistance of 0 ohms.

VT = 3 Volts
RT = 3k Ohms
IT = 3/3k = 1mA
I1 = 1mA
I2 = 1mA
V1 = 1V
V2 = 2V

NOTE: The current flows clockwise.


Finally:

The combination of VS1 + VS2 results as:
RT = 3k Ohms
I1 = ILVS2 - ILVS1 = 500uA
I2 = I2VS2 - I2VS1 = 500uA
V1 = V1VS2 - V1VS1 = 500mV
V2 = V2VS2 - V2VS1 = 1V

Total voltage drops = 1.5V

NOTE: I am subtracted the lower values from the higher values since the current is flowing in opposing directions.

When I measured these points in the schematic with real resistors and power supplies in class, I measured just under 2 Volts (resistors at 5% tolerance). When I measure these points with this exact schematic in Multisim I measure 2 volts exactly. I still can't figure out how I'm supposed to calculate 2 volts?

 

[JohnRoberts]

Using Superposition Theorem:

VS1 becomes R1 + R2 in series with 1.5v source. The 3 volt source is replaced with its internal resistance of 0 ohms.

VT = 3 Volts
RT = 3k Ohms
IT = 1.5/3k = 500uA
I1 = 500uA
I2 = 500uA
V1 = 500mV
V2 = 1V

NOTE: The current flowers counter-clockwise

VS2 becomes R1 + R2 in series with 3v source. The 1.5 volt source is replaced with its internal resistance of 0 ohms.

VT = 3 Volts
RT = 3k Ohms
IT = 3/3k = 1mA
I1 = 1mA
I2 = 1mA
V1 = 1V
V2 = 2V

NOTE: The current flows clockwise.


Finally:

The combination of VS1 + VS2 results as:
RT = 3k Ohms
I1 = ILVS2 - ILVS1 = 500uA
I2 = I2VS2 - I2VS1 = 500uA
V1 = V1VS2 - V1VS1 = 500mV
V2 = V2VS2 - V2VS1 = 1V

Total voltage drops = 1.5V

yup, 1.5v is total drop between 3V and 1.5V

NOTE: I am subtracted the lower values from the higher values since the current is flowing in opposing directions.

When I measured these points in the schematic with real resistors and power supplies in class, I measured just under 2 Volts (resistors at 5% tolerance). When I measure these points with this exact schematic in Multisim I measure 2 volts exactly. I still can't figure out how I'm supposed to calculate 2 volts?

I'd claim that I am rusty with this, but that would mean I once knew it.  ;D

Your calculation of the current is correct and direction. So to determine the final voltage at junction of R1 and R2 that 500uA times either 1K or 2k tells you voltage difference above the 1.5V (1.5V + (1k x 500uA)) or below the 3V ( 3v-(2k x 500uA)).

I find it a lot easier to just look at it.. but don't let me interfere with your proper book larnin...

JR
 

 

[ruffrecords]

For me, the secret is to combine the two batteries into one and this is easiest if you convert them first to Norton equivalent circuits i.e. a current source in parallel with a resistor. So, the 1.5V battery and 1K become a 1.5mA current source in parallel with 1K. The 3V battery and and 2K resistor become a 1.5mA source in parallel with 2K. To combine these into one, add the currents and work out the parallel resistance, so it become a 3mA source and a 2/3 K ohm resistor. Convert that back the Thevenin and you get a voltage of  3 * 2/3 = 2V and a resistance of 2/3 K ohms.

Then this in series with 470 + 510 is 1646 ohms which means the current thru the 510 is 3/1.646 = 1.822mA and the volts across the 510 is thus 929mV.

Cheers

Ian

 

[audiomixer]

When I measured these points in the schematic with real resistors and power supplies in class, I measured just under 2 Volts (resistors at 5% tolerance). When I measure these points with this exact schematic in Multisim I measure 2 volts exactly. I still can't figure out how I'm supposed to calculate 2 volts?

when you look at the unloaded situation you have 1k + 2k across 1.5V (difference between 3V and 1.5V source, both referenced to 'ground'.... now the voltage drop is going to be distributed across those resistors according to their ratio. R2/R1 = Vr2/Vr1. the 1.5V thru 3k total will split int 0.5V and 1V respectively and bring the voltage at that point to 2V as from both sources of course, they cannot be different ;-) there would be some current (1.5/3k) flowing into your 1.5V battery.

I hope this helps you understand more ....

- michael

 

[sahib]

Barking up the wrong theorem tree unfortunately. And it is very easy to fall into that trap.

One end of each resistor is connected to the voltage sources. The other ends of the resistors are joined and formed say node A.

The other ends of the voltage sources also joined together and formed say node B.

Then both voltage sources  in parallel hence Millman's Equivalent Voltage Theorem applies.


Veq = (1.5/1000) + (3/2000) / (1/1000) + (1/2000)

= 0.0015 + 0.0015 / 0.0015

= 0.0030 / 0.0015

= 2V

 

[moamps]

The answer is supposed to be 2 volts, but I can't figure it out... I calculate Thevenin's resistance as 1137 Ohms which I'm confident is correct. Could anyone walk me through calculating VTh?

Vth=3V-Iloop*R2

Iloop=(3-1.5)/(R1+R2)=1.5/3=0.5mA

Vth=3v-0.5mA*2000=2V

V(Rl)=Vth*Rl/(Rth+Rl)=2*510/(1137+510)=0.62V

 

[moamps]

..... Convert that back the Thevenin and you get a voltage of  3 * 2/3 = 2V and a resistance of 2/3 K ohms.

Then this in series with 470 + 510 is 1646 ohms which means the current thru the 510 is 3/1.646 = 1.822mA and the volts across the 510 is thus 929mV.

IMO,

I=2/1.646

Regards,
Milan

 

[JohnRoberts]

;D ;D ;D

classic internet...  no shortage of answers....  pick one.

JR

 

[Matador]

Thevenin's theorem says you can find the equivalent circuit by doing two steps:
1) Open the circuit between the two terminals and find the voltage across them
2) Short across the two terminals and find the current

So for this circuit, if you open up across AB, the 510R load resistor and the 470R resistor drop away, as there is no current through them.

So you're left with a single loop KVL.  Taking this loop equation (assuming the current is called I1, and that the current goes clockwise):

1.5V - I1*R1 - I1*R2 - 3V = 0

This is a single equation in a single unknown, so you can solve for I1.  This yields an answer of -0.5mA.

This means that the current is going "up" (or counter-clockwise), which means there is a voltage drop across R2!  Ohms law says that 0.5mA across 2K is a 1V drop.  So the voltage at A is the 3V supply - 1V drop = 2V.

You can also arrive through the other side.  The 1.5V supply would be added to the drop across R1, which is 1K*0.5mA = 0.5V, and this is a voltage rise, so the voltage at A is 1.5 + 0.5 = 2V.

To find the Thevenin resistance, you short across those terminals and find the short circuit current.  So the 510R resistor is again lost.  This yields two KVL loop equations to find the current, and then you can find the equivalent resistance by dividing this current by the 2V you found above.

Hope this is simple enough.  There are at least three or four ways to skin this cat.

 

[PRR]

Thevenin is general and covers ALL topologies.

My tiny mind (also SPICE's) prefers to mark one node "zero" and work around that, if possible.

This problem, both batteries and one result-point are common. It seems obvious to call that "zero" to simplify notation. However we could call any other point, or no point, "zero". However our tiny mind may fog-up.

When faced with a foggy problem, clean-up, simplify, then build-back.

First crack: the R1-R2 node is between 3V and 1.5V, and closer to the 1.5 than to the 3.

Further, if the 510 has been tossed-out, then R3 has no effect on point A voltage.

 

[moamps]

;D ;D ;D
classic internet...  no shortage of answers....  pick one.
JR

And SNR tends to zero .....

 

[CJ]

ok

 

[CJ]

here is the standard RDH4 approach to the whole network, not the easy problem with the 510 resistor ommited which had an answer of 2 volts..

the circuit was redrawn as shown in the pic

current direction was assigned according to intuition,

if you do not know which way, simple assign directions and the solution will still work out, if you get a negative answer, that means that the current goes the opposite way that you picked.

we have combined the 510 and 470 into 980 ohms.

all currents pointing to node "b" will be given a positive value,

equations-

i fb=i bc + i ab

circuit dabcd:      +3 -2000 i ab -980 i bc = 0

circuit dabfgcd:    +3 - 2000 i ab + 1000 i fb = 0

circuit fbcgf:        +1.5 - 1000 i fb -980 i bc = 0

arrange i fb = i bc + i ab : 

i ab = i bc - i fb

sub into circuit dabcd:

+3 -2000(i bc -i fb) -980 i bc = 0

simplify:

+3 - 2980 i bc + 2000 i fb = 0


now pick a circuit equation with the same 2  currents and solve simul"

+3 - 2980 i bc + 2000 i fb = 0  and

+1.5 - 1000 i fb - 980 i bc = 0

we can multiply both sides of the bottom eq by 2, so we have

+3 - 2980 i bc + 2000 i fb = 0  and

+3 - 2000 i fb -1960 i bc = 0
add the two equations and we get:

6 - 4940 i bc = 0,  i bc = 6/4940,

i bc = 0.001215 amps

plug this into

+3 -2000 i ab - 980 i bc = 0 and we get

+3 - 2000 i ab - 980 (0.001215) = 0

i ab = 3 - 1.1907 - 2000 i ab = 0

i ab = 1.8093/2000 = 0.000905 Amps

all we need is i fb,

well i fb = i bc - i ab, so

i fb = 0.001215 - 0.000905 = 0.00031 Amps, so

i ab = .905 ma
i bc = 1.215 ma
i fb = 0.31 ma

so the voltage at node "b" will be

0.001215 * 980 = 1.1907 volts dc.

i messed this up the first time but that was to scare away the guy who was cheating on his homework, jus kiddin, 

this is straight out of RDH4, page 130.