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