Hi
kirchhoff's law

I can do first loop in circuit but in second loop I not understand how I can do loop!
can you help me and explain how I can do second loop .

this is not homework,

thanks

 

loop1:
KVL
10V - 2I1 - 6I3 - 4I1 =0

loop2:

-15V + 8I2 + 6I3 =0

why source is negative in loop2?
and resistors are positive in loop2?
thanks,

 

I3 = I1 + I2

10 = 2I1 + 6I3 +4I1 = 2I1 + 6I1 + 6I2 + 4I1 = 12I1 + 6I2

15 = 6I3 + 8I2 = 6I1 + 6I2 + 8I2 = 6I1 + 14I2

Can you solve it from there?

 

I3 = I1 + I2

10 = 2I1 + 6I3 +4I1 = 2I1 + 6I1 + 6I2 + 4I1 = 12I1 + 6I2

15 = 6I3 + 8I2 = 6I1 + 6I2 + 8I2 = 6I1 + 14I2

Can you solve it from there?

can you explain how you are do loop 2 (kirchhoff's rules) ?
15 = 6I3 + 8I2
thanks

 

why this circuit is different for first circuit

in first circuit the loop 2 is {-15V + 8I2 + 6I3 =0}
but in second circuit the loop2 is { -2 + 8 +3Ix =0}

please can you explain what differences ?!

thanks ,

 

 

There are many methods to write KVL.
For example in first circuit for the voltage sources if you go from "plus to minus" you add " -" and if you go from "minus to plus" you have "+".
And for the resistors. If the resistor current is in the same direction as loop. We give "-", but if current flow in opposite direction the the loop, we give "+".

First circuit loop 1
We start at the bottom of a 10V voltage source (point E). And we go from minus to plus so we have +10V. Next we go from point A to B. I1 is in the same direction as our loop so we give "-". +10V - I1*2. Next we go from B to D. And again I3 is in the same direction as loop, so again "-".
+10V - I1*2 - I3*6. And we have the same situation for 4 ohm resistor.
And this is why we end up wit this
10V - I1*2 - I3*6 - I1*4 = 0

Now the second loop.
We start at point B and we go from B to C (from "plus to minus" ), so we add -15V.
Next from C to D. Notice that now I2 and I3 currents opposite our loop "current", so must give "+". So finally we end with this. -15V + I2*8 + I3*6 = 0

For the second circuit the book use different method (convention).
They first label the voltage drops across resistors.
Loop 1



Next they start at point A and go to point B. At point A we firs we see a "minus", so we write
-12V. As we go from B to C , at point B we see "plus", so -12V +v;. And form C to A we also see "plus".
And this is why we have
-12V + v + 2V = 0

For the second loop we again start at A and we go from point A to point C. And first what we see At point A is a "minus" sing -2V.
At point C we see "plus" -2v +8V and at point D we also see "plus".
So we end up with -2V + 8V + 3Ix = 0

We can use this method to analysis the first circuit.
Current flow from "+" to "-" so we label the voltage drops based on the direction of currents. In resistors current must enter the positive voltage polarity node.



Loop 1 - start at point A

I1*2 + I3*6 + I1*4 - 10 =0

Loop 2 - start at D

-I3*6 + 15 - I2*8 =0

And sorry for my poor English

 

thank you very much sir, this is very kind of you ...

 

I use 2 methods in this circuit but the result not same
I think the first method is true and second method is wrong
please sir can you explain how I knew currents direction in any circuits?


method 1:
-15 +1I +2Vx +5I + 2I=0



method 2:
15 - 1I -2Vx +5I + 2I=0

sorry for my poor English

thank you,

 

Generally, it doesn't matter. Assume a direction. Since voltage drops each carry a +- sign, if you guess wrong, the algebra will take care of it.

btw- I think this problem is a trick question. I compute the current to be zero.

 

I use 2 methods in this circuit but the result not same
I think the first method is true and second method is wrong
please sir can you explain how I knew currents direction in any circuits?
View attachment 72646,

No, in this circuit you have only one loop.
And to avoid confusion try to stick to one "sign convention"

http://hades.mech.northwestern.edu/index.php/Kirchhoff's_Current_and_Voltage_Laws#Kirchhoff.27s_Voltage_Law_and_Loop_Analysis
http://web.engr.oregonstate.edu/~traylor/ece112/lectures/kvl.pdf
http://www.niu.edu/iteams/documents/ueet602/KVL and KCL.pdf

btw- I think this problem is a trick question. I compute the current to be zero.

Very strange because I get I = 5/6A = 0.833A

 

...
Very strange because I get I = 5/6A = 0.833A

Jony,

Do you agree that if I = 5/6, then V(x)-V(y) = I/2 = (5/6)/2 = 0.416667V?

 

Jony,

Do you agree that if I = 5/6, then V(x)-V(y) = I/2 = (5/6)/2 = 0.416667V?

Isn't 2Vx is a voltage controlled voltage source?

 

Isn't 2Vx is a voltage controlled voltage source?

I took it as being a voltage-controlled current-source with a transconductance of 2. That is why I got I = 0A.

If I take it as a voltage-controlled voltage-source, then I get 0.83333A flowing from X to Y.