# calculating currents using KCL and KVL

Discussion in 'Electronics Homework Help' started by Joe Bachi, Feb 12, 2013.

1. ### Joe Bachi

21
0
Feb 3, 2013

if i(a)= 4A and i(b)= -2A
What is the current value of the current source?
Any tips or suggestions?
Thanks

2. ### Joe Bachi

21
0
Feb 3, 2013
using KVL in the top mesh:
10i(a) + 11i(a) + 9i(a) - 5[i(b)-i(a)] - 30[i(g)-i(b)+i(a)]
i(g) = -1 A
Would that make sense?

3. ### Laplace

1,252
184
Apr 4, 2010
To write the node equations for this circuit I added a reference ground to the minus terminal of the power supply and combined all series resistors, then put labels the three nodes as shown.

Solving the node equations for the given constraint conditions leads to an inconsistency for the value of current source 'I'. Don't believe any solution is possible.

File size:
15.9 KB
Views:
199
File size:
115.8 KB
Views:
278
4. ### Harald KappModeratorModerator

11,266
2,577
Nov 17, 2011
I have come to the same result.
At least one of the initial conditions is wrong.

5. ### Joe Bachi

21
0
Feb 3, 2013
What if i(a)=-2 and i(b)=4?
Would we be able to get a result then?
In the textbook they didn't really specify which was i(a) and which was i(b)

(This is my first electrical engineering course and we were told to only use KVL, KCL, or Ohm's law in order to solve this)
And BTW Thanks for the time you took writing those replies

6. ### Harald KappModeratorModerator

11,266
2,577
Nov 17, 2011
Do you have a better copy of the image?
The directions of the arrows are almost indiscernible. The top arrow, for example, could also point towards the right.
Also in you manual drawing the bottom right resistor is 15 Ohm whereas in the copy it looks like 16 Ohm.

The way your reasoning goes is o.k. (apart from the direction and the sign of the current source, which I would have backwards due to the direction of the arrow in the current source, but that's a minor aside).
You can cross check your result by evaluating the two other loops (using the values you calculated from the top loop) and checking for consistency of the values you get.

7. ### Laplace

1,252
184
Apr 4, 2010
Indeed, i.a is the current at the lower left corner (flowing to the right) while i.b is the current at the upper right corner (flowing to the left). The first step is to use i.a and i.b to calculate voltage drops, then use KVL around the outer loop to get the voltage across the 16+4 resistor, and from that the current. Then use KCL at the node with the '+' terminal of the voltage source to find the current through the 5 ohm resistor. Yes, the current is flowing in the reverse direction through the voltage source. All the voltage and current values are whole numbers. I calculated 190 volts across the current source and verified KCL at all nodes. Even the power calculation worked out OK (with 400 watts being dissipated in the voltage source).

8. ### Joe Bachi

21
0
Feb 3, 2013
Yes, that seems logical,
Thank you very much
(I'm loving this forum! hopefully with time I'll be the one answering a few questions)