Ok I am getting this weird value for resistance.. 108.33 ohm which i am pretty sure it's wrong

So please help me out with it. I know that 1st part V=-14750 is correct
But for finding resitance, i don't know what's wrong.

I have attached my work and the question.r

 

It would help if you would describe what the problem is. If I hadn't looked at your worksheets, I wouldn't know that apparently, you're supposed to remove the 25 ohm resistor and determine the Thevenin equivalent of what's left. Is that correct?

 

Yes. I removed the 25 ohm resistor and calculated the voltage Vab.

Then I put the 1 ampere source between A and B since there is a dependent source in the circuit and I short cktd the 5 Ampere current source. I again applied KCL at both A and B. I got weird answer for Vab and Rab is supposed to be Rab = Vab/1

 

You have correctly realized that Vth is the voltage between Va and Vb; a differential voltage, in other words. Often, Vth would be between some terminal and ground, but not in this case.

So, Rth is also a differential impedance.

With slight rearrangement, your first equation is:

Va/50 = 5

and the second is;

Vb/100 = 150

When you solved these, you got Va = 250 and Vb = 15000. Vth is the difference between these; Vth = 250-15000 = -14750.

Ordinarily, the 3V source would be considered a dependent current source, but in this case, since the 5A source is connected in series with the 10Ω resistor, the current through the 10Ω is constant; it doesn't depend on the voltage at Va. So, the 3V source is really an independent source.

To find Rth, we need to connect a 1 amp current source between Va and Vb, with the independent current sources reduced to open circuits. If a 1 amp current source is connected between Va and Vb, with the current entering the Va node, and leaving the Vb node, your equations become:

Va/50 = 1

Vb/100 = -1

If we solve this system, we get Va = 50 and Vb = -100.

The differential impedance is equal in value to the difference in the two voltages Va and Vb. This is just ohm's law; if you pass 1 amp through an unknown resistance, the voltage across that resistance is equal in value to its resistance.

Thus, Rth is:

Rth = Va-Vb = 50-(-100) = 150Ω

It's actually easier than this; you can determine Rth by inspection.

If you remove the current sources and the 25Ω resistor, what's left? The 50Ω resistor and the 100Ω resistor, connected together at the bottom and open at the top with the Va node being the top of the 50Ω and the Vb node being the top of the 100Ω. So, if you connected an ohmmeter between the remaining Va and Vb, what would it read? It would read 150Ω

 

I got what you are saying but then when i did this circuit in muiltism program, and calculated the current it said -98.334

But if you calculate current with these values, -14750 / 175 = 84.28 A

What's wrong?

 

What voltage did you get at the junction of the 10 ohm, 25 ohm and 50 ohm?

hgmjr

 

-14750 volts

 

-14750 volts

Actually that is the value you obtained for Vth. I am referring to the voltage that you obtain at the junction of the 10, 25, and 50 ohm resistor once you reconnect 25 ohm resistor.

hgmjr

 

-2.107 Kilovolts

 

Can you draw the original circuit and label the voltages and currents that you have calculated? That way we can see which, if any, values need additional attention.

hgmjr

 

I didn't calculate. I actually checked it in a program on my laptop. But anyway, is the current -84.28 correct or -98.334?

 

-84.28 should be correct..?

 

I got what you are saying but then when i did this circuit in muiltism program, and calculated the current it said -98.334

But if you calculate current with these values, -14750 / 175 = 84.28 A

What's wrong?

Why did you divide by 175? The value of Rth as I showed in post #4 is 150.

If you calculate -14750/150 you get -98.33333.

 

Because we connect the thevenin circuit with 25 ohm resistor right!
And applying kvl, 14750 + 25I + 150I=0 That's how.. ??

 

The Thevenin equivalent is a source of -14750 volts in series with a 150 ohm resistor. If you want to calculate the short circuit current, Isc, of the Thevinin equivalent, it's just -14750/150 = -98.33333 amps.

If you want the current in a 25 ohm resistor connected to the a-b terminals, then it will be -14750/175 = -84.2857 amps.

In post #5, you didn't say which current you're talking about.

In post #11 you are also not clear which current you're asking about.

If you connect a 25 ohm resistor to the Thevenin equivalent, then you will have a voltage divider and the voltage across the 25 ohm resistor will be given by:

-14750 * 25/(25+150) = -2107.142857 volts

Which current are you determining with Multisim? The current a 25 ohm resistor connected to a-b, or the short circuit current in a short applied to the a-b terminals?

 

The Thevenin equivalent is a source of -14750 volts in series with a 150 ohm resistor. If you want to calculate the short circuit current, Isc, of the Thevinin equivalent, it's just -14750/150 = -98.33333 amps.

If you want the current in a 25 ohm resistor connected to the a-b terminals, then it will be -14750/175 = -84.2857 amps.

In post #5, you didn't say which current you're talking about.

In post #11 you are also not clear which current you're asking about.

If you connect a 25 ohm resistor to the Thevenin equivalent, then you will have a voltage divider and the voltage across the 25 ohm resistor will be given by:

-14750 * 25/(25+150) = -2107.142857 volts

Which current are you determining with Multisim? The current a 25 ohm resistor connected to a-b, or the short circuit current in a short applied to the a-b terminals?

You said that The Thevenin equivalent is a source of -14750 volts in series with a 150 ohm resistor. If you want to calculate the short circuit current, Isc, of the Thevinin equivalent, it's just -14750/150 = -98.33333 amps.

But apparently when you connect 150 and 25 ohm in series with -14750 V, the current in the circuit is -14750/175 = -84.2857 amps. so this means this is the current in 150 ohm and 25ohm resistor also both

 

You said that The Thevenin equivalent is a source of -14750 volts in series with a 150 ohm resistor. If you want to calculate the short circuit current, Isc, of the Thevinin equivalent, it's just -14750/150 = -98.33333 amps.

But apparently when you connect 150 and 25 ohm in series with -14750 V, the current in the circuit is -14750/175 = -84.2857 amps. so this means this is the current in 150 ohm and 25ohm resistor also both

Since the 150 Rth and the added 25 ohm resistor are in series, the current in the series combination will be -84.2857; that will be the current in each of them. Is this what you're asking?

 

Yes that's what i am asking. So it should be -84.28 right? in each one of them.

Is it correct or not?

 

I repeat:

Since the 150 Rth and the added 25 ohm resistor are in series, the current in the series combination will be -84.2857; that will be the current in each of them.

This is how a Thevenin equivalent works, and how a series circuit works. When two resistors are in series, the current (if there is any) in both of them is the same.

 

I get what you are saying. Thanks a lot!