# Applying Ohm's and Kirchhoff's Laws Circuit

Discussion in 'Homework Help' started by kinetickabuki, Oct 20, 2014.

1. ### kinetickabuki Thread Starter New Member

Oct 20, 2014
1
0
I must be doing something wrong. I've tried using KCL, KVL, Ohm's law, node analysis, and substitution to figure this out to no avail. Here's what I've tried:

i(7)=V(1)/70 kohms
i(2)=V(2)/20 kohms
i(3)=V(3)/30 kohms
V(3)=i(3)(30 kohms)
V(I)=5mA(R)

10V=-i(7)(70 kohms)+V(3)-V(I)+i(2)(20 kohms)

i(7)+i(2)-i(3)+5mA=0

Substitution with loops:
i(7)=[-10-30 kohms(i(3))]/-70
[-10-30 kohms(i(3))]/-70=i(3)-i(2)-5mA

Any suggestions?

File size:
46.5 KB
Views:
61
2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
KVL Left Loop: 10 = i7*70k + i3*30k
KCL Current through the 30k: i3 = i7 + i2
By inspection: i2 = 5mA

Substitution: 10 = i7*70k + 30k*(i7 +5m) = 100k*i7 + 150
Rearrange: i7 = -140/100k =-0.14mA (correction: -1.4mA)

Every thing else follows...

Confirmation:

Last edited: Oct 21, 2014

Feb 17, 2009
3,990
1,115
4. ### WBahn Moderator

Mar 31, 2012
18,077
4,907
Instead of just feeding you answers, I will try to explain where you are going wrong. Your basic problem, which is very common, is that you don't appear to truly understand Ohm's Law or KVL, KCL, though it's hard to tell because you use variable that you don't define and that aren't indicated on your diagram. What are V(1), V(2)?

Remember that Ohm's Law relates the voltage ACROSS a resistor and the current through THAT resistor to the resistance of THAT resistor. You can't use a voltage in Ohm's Law unless it is the voltage across that particular resistor. Also, voltages and currents have polarities that can't be ignored. Even if I assume that what you mean by V(1) and V(2) are the voltages across the 70k and 20k resistors respectively, you don't define what their polarities are. ALWAYS clearly define your variables -- drawing them on the diagram is the best approach, but describing them in words also works. Don't make people guess.

KVL is merely the process of summing up the voltage gains/drops across each element as you go around a closed loop. Look at your KVL equation. You have terms in there that don't all belong to the same loop. When doing KVL, go element by element and add terms that represent the voltage gain/drop across each element as you proceed to cross each element as you go around a closed loop. If you don't cross an element, then don't include a voltage for it. The safest way, until you get comfortable and competent enough to take shortcuts, is to pick a point on the circuit and decide if you are going to add voltage gains or voltage drops. Then proceed around the circuit adding up whichever you chose until you get back to the point you started at and set the result equal to zero.

KCL is merely the process of saying that any current that enters a point must leave that point. Your KCL equation is fine until you tack on that 5mA. The current in the current source is not (directly) going to the top center node, which is the point you are applying KCL to. If there had been an i10 shown going through the voltage source, would you have tacked that onto your equation, too?

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,390
1,211
Mike,

Your calculation states I(7) is -0.14 mA. Your simulation states it's -0.0014 amperes. I'm sure you had a typo somewhere.

6. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Any idiot should be able to see that 140/100k = 1.4mA ( I'm the idiot). In my defense, look at the time stamp of when I posted...

7. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,390
1,211
I knew it was a typo .... and yes, we are all susceptible to those errors at that hour.