Engineering Is my answer wrong? simple circuit problem, 10.4 mA

[PainterGuy]

Hello everyone,

I solved the circuit diagram in the following link for I₁, I₂, I₃.

http://img59.imageshack.us/img59/58/circuitchap8.jpg

I took I₃=I₁+I₂. My value for I₃ was 10.4 mA or 0.0104 A. The answer given in the book is:

I₁ : CW, 1.445 mA
I₂: CCW, 8.513 mA
I₃: Down, 9.958 mA

I don't know if my answer is incorrect or the difference is because of decimal places. The difference between book's answer and mine's is 0.442. Is my answer wrong? Help me out with this please. Much grateful.

Cheers

 

[vk6kro]

I get:

1.44505 mA
-9.95796 mA
8.51287 mA

So, the book is right.

The easy way to check is that the voltage at the top of the 2.2 K resistor is 1.90751 volts and the result of taking the voltage across each resistor away from each power source should equal this for the 5.6 K and 3.3 K resistors.

The middle resistor has 21.90751 volts across it (2.2 K * 9.95796 mA). Subtract the 20 volt power source to get 1.90751 volts.

LTSpice is excellent for checking stuff like this. You should get a copy of it.

 

[PainterGuy]

Hello everyone,

I solved the circuit diagram in the following link for I₁, I₂, I₃.

http://img59.imageshack.us/img59/58/circuitchap8.jpg

I took I₃=I₁+I₂. My value for I₃ was 10.4 mA or 0.0104 A. The answer given in the book is:

I₁ : CW, 1.445 mA
I₂: CCW, 8.513 mA
I₃: Down, 9.958 mA

I don't know if my answer is incorrect or the difference is because of decimal places. The difference between book's answer and mine's is 0.442. Is my answer wrong? Help me out with this please. Much grateful.

Cheers

I get:

1.44505 mA
-9.95796 mA
8.51287 mA

So, the book is right.

The easy way to check is that the voltage at the top of the 2.2 K resistor is 1.90751 volts and the result of taking the voltage across each resistor away from each power source should equal this for the 5.6 K and 3.3 K resistors.

The middle resistor has 21.90751 volts across it (2.2 K * 9.95796 mA). Subtract the 20 volt power source to get 1.90751 volts.

LTSpice is excellent for checking stuff like this. You should get a copy of it.

Many thanks vk6kro. Much obliged for your help.

But my value for I₃ is also close to the book's for I₃. I see your value -9.95796 mA has -ve sign. Is it I₂? If I take the current running CW +ve then I₁ would be +ve and I₂ negative because it's running CCW. I₃ is going downward which means it's also CW. Is there any chance that my value for I₃ of 10.4 mA to be correct? Tell me please.

And thanks for telling me about that simulation software LTspice. I see MultiSim and CircuitLogix are also popular. Which software of these three is easy to use.

Cheers

 

[vk6kro]

Is there any chance that my value for I3 of 10.4 mA to be correct? Tell me please.

No, it should be 9.95796 mA. 10.4 mA is nowhere near this value. I think you should check your equations.

And thanks for telling me about that simulation software LTspice. I see MultiSim and CircuitLogix are also popular. Which software of these three is easy to use.

I have mostly used LTSpice 4. I find it excellent and simple to use.
It can be used at a much higher level than simple sircuits like this.
Just the circuit drawing part of the software is amazing.
It is also free.

I used Multisim a bit, but I seemed to spend a lot of time trying to get the simulator working properly and not really getting practical results.

 

[PainterGuy]

Many thanks vk6kro. I would try to solve it again and if got different answers then would post the equations here.

I have also download LTspice's trial version. Will start playing around with it soon.

In one of the sections in the book it specifically use the term "independent" loops while using KVL. For example, have a see on these pages (highlighted parts):
https://docs.google.com/viewer?a=v&...3MjgtNTQ5YzJjNDQ4OTc2&hl=en&authkey=CMe2mfwD"
https://docs.google.com/viewer?a=v&...lMTAtNTlhYWRmZDE1MGNm&hl=en&authkey=CP79wKgE"

Could you help me to know the difference between "independent" loops and "normal" loops please? As you can see on Page 2, there can be three loops instead of two. The books doesn't care about the outer loop which will include E₁ and E₂. Help me out please. Much grateful.

Cheers

 

[The Electrician]

Could you help me to know the difference between "independent" loops and "normal" loops please? As you can see on Page 2, there can be three loops instead of two. The books doesn't care about the outer loop which will include E₁ and E₂. Help me out please. Much grateful.

Cheers

As you have noted, your circuit could have 3 loops, but only 2 loops are needed to solve the circuit. If you write 3 loop equations, one of the equations will be redundant; it will not give any "new" information in the solution. The impedance matrix will be singular, and there will not be a solution. But, if you pick any 2 of the 3 possible loops they will be "independent" and the system of equations will have a solution.

 

[PainterGuy]

As you have noted, your circuit could have 3 loops, but only 2 loops are needed to solve the circuit. If you write 3 loop equations, one of the equations will be redundant; it will not give any "new" information in the solution. The impedance matrix will be singular, and there will not be a solution. But, if you pick any 2 of the 3 possible loops they will be "independent" and the system of equations will have a solution.

Many thanks The Electrician.

So including the outer loop, the third one, which will include E1 and E2 is a bad thinking because firstly it wouldn't involve any new components such as resistors. All the components have already been included in the two loops used by the book. So, using third one is also a nonsense. I have a choice either I use two loops used in the book or include the outer loop and one of the two loops used by the book.
In other words any loop is independent if it introduces a new component(s) which has not been covered in any other loops.

I am grateful for you help.

Cheers