Hello all
I applied the loop rules to the circuit by solving loops although it is a simplest 2 loop circuit but the difficulty I am facing is in calculating 'Vout'. I1 & I2 are -0.05mA & 1.3mA respectively. Did I calculated the values precisely.??
Please if someone can reply with calculations of this circuit leading to finding Vo, his help will be much appreciated.
Thanks

 

No, your answer is wrong. Show us your nodal equations.
And the correct answer is I1 = -0.1mA and I2 = 1.1mA

 

No, your answer is wrong. Show us your nodal equations.
And the correct answer is I1 = -0.1mA and I2 = 1.1mA

Well thanks first of all. here are the equations:
For Loop 1:-
2I1 + 4I1 + 2(I1-I2) + 3 = 0 ===>> 8I1 - 2I2 = -3 ...................(1)
For Loop 2:-
-3 + 2(I2-I1) - 6 + 6I2 = 0 ===>> -2I1 + 8I2 = 9 ....................(2)

yeah you are right I rechecked the calculations I2 = 1.1mA but I1 = -0.5mA.
But you didn't present the solution for Vo even with the values you told. Sir

 

Well thanks first of all. here are the equations:
For Loop 1:-
2I1 + 4I1 + 2(I1-I2) + 3 = 0 ===>> 8I1 - 2I2 = -3 ...................(1)
For Loop 2:-
-3 + 2(I2-I1) - 6 + 6I2 = 0 ===>> -2I1 + 8I2 = 9 ....................(2)

yeah you are right I rechecked the calculations I2 = 1.1mA but I1 = -0.5mA.

Well, please recheck your calculations again.
Here you can find the solution
http://www.wolframalpha.com/input/?...I_2-I_1)+-+6+++6I_2+=+0+solve+for+I_1+and+I_2

But you didn't present the solution for Vo even with the values you told. Sir

Simply use the ohm's law to find Vo. Which current is flow through 6k resistor?

 

Well, please recheck your calculations again.
Here you can find the solution
http://www.wolframalpha.com/input/?i=2I_1 + 4I_1 + 2(I_1-I_2) + 3 = 0 , -3 + 2(I_2-I_1) - 6 + 6I_2 = 0 solve for I_1 and I_2

Simply use the ohm's law to find Vo. Which current is flow through 6k resistor?

I got here to just close this thread which I posted in what a dull mood I was while solving this easiest circuit now after some time I just tried it again before coming to see new reply (solved myself already) and i was shocked how I was making easy things complex by assuming it difficult to solve (I was out of energy after practicing a lot of circuits ). I feel shame Sorrrrrrrrrrryyyyyyyyyy.
thanks jony130

 

Now How do I close this thread.??????

 

Well thanks first of all. here are the equations:
For Loop 1:-
2I1 + 4I1 + 2(I1-I2) + 3 = 0 ===>> 8I1 - 2I2 = -3 ...................(1)
For Loop 2:-
-3 + 2(I2-I1) - 6 + 6I2 = 0 ===>> -2I1 + 8I2 = 9 ....................(2)

yeah you are right I rechecked the calculations I2 = 1.1mA but I1 = -0.5mA.

We aren't mind readers. You never defined I1 and I2, so not only do we not know which mesh is associated with which current, we don't know what direction either of them is going. Don't make people guess at what you mean -- particularly people for whom you are asking free assistance from. Engineering is not about guessing.

Have you checked your answers?

Does (8 kΩ)(1.1 mA) - (2 kΩ)(-0.5 mA) = -3 V ?
Does -(2 kΩ)(1.1 mA) + (8 kΩ)(-0.5 mA) = 9 V ?

Since the second term in both equations is an integer and since the right hand side is an integer, what is required to be true of the first term in both cases? Is the requirement met?

Always, always, ALWAYS ask if your answers make sense.

You also need to track your units throughout your work. Most mistakes you make will mess up the units and you can catch them real quick -- if the units are there to get messed up.

But you didn't present the solution for Vo even with the values you told. Sir

Good. It's YOUR homework. You are expected to show YOUR attempt first.

 

Now How do I close this thread.??????

You don't.

It is not uncommon for someone else to come upon a thread that the OP is done with and to have some questions about it and to revive it for their own benefit. The whole idea of the forum is for lots of people to benefit from each person's struggles and experiences.

And no need to feel ashamed about anything. You are learning. And even after you have this stuff down pat there will be days that you will make the most obvious mistake and spend hours looking at it and right past your mistake because you mind knows what you meant to do and will refuse to see what you actually did.

 

We aren't mind readers. You never defined I1 and I2, so not only do we not know which mesh is associated with which current, we don't know what direction either of them is going. Don't make people guess at what you mean -- particularly people for whom you are asking free assistance from. Engineering is not about guessing.

Have you checked your answers?

Does (8 kΩ)(1.1 mA) - (2 kΩ)(-0.5 mA) = -3 V ?
Does -(2 kΩ)(1.1 mA) + (8 kΩ)(-0.5 mA) = 9 V ?

Since the second term in both equations is an integer and since the right hand side is an integer, what is required to be true of the first term in both cases? Is the requirement met?

Always, always, ALWAYS ask if your answers make sense.

You also need to track your units throughout your work. Most mistakes you make will mess up the units and you can catch them real quick -- if the units are there to get messed up.



Good. It's YOUR homework. You are expected to show YOUR attempt first.

O hey brother cool down u started without brakes. I didn't ask the thing which does not make SENSE I have posted a schematic diagram in my question if you got a second to watch out what i am asking for. see the top of this page where i posted my query, there are all parameters available which are needed to solve the circuit.

 

O hey brother cool down u started without brakes. I didn't ask the thing which does not make SENSE I have posted a schematic diagram in my question if you got a second to watch out what i am asking for. see the top of this page where i posted my query, there are all parameters available which are needed to solve the circuit.

Really? Okay, perhaps I'm missing something. Where do you define what the currents I1 and I2, including direction, are?

You gave answers which are not solutions to the equations you gave. Did you bother to check to see if they were?

I'm trying to help you improve your work so that you get the correct answers more often and, more importantly, will catch most of the time when you have incorrect answers. But if that's not important to you....

 

well WBahn I appreciate your effort. Thanks

 

Azalan, without giving too much help..I suggest merging the 4k Ohm, 2 Ohms series resistors together. You can redraw the circuit to make it look different. The voltage drop across the 6kOhm resistor due to the 3v is going to be found with a voltage divider equation.... You need to take 3v and do some manipulation. Won't say anymore because I don't want my head bitten off. But ahem.. I recommend voltage divider equation.

 

I have came up with the following Mesh equations:

M1:
2kI1 + 4kI1 +2kI1 -2kI2 + 3 = 0 => 8kI1 -2kI2 = -3

M2:
-3 + 2kI2 -2kI1 -6 +6kI2 =0 => -2kI1+8kI2 = 9

Solving the system above gives:

I1= -0,1mA
I2= 1,1mA

Thus, V0=6k*I2 = 6.6V