# Mystified by current in a circuit diagram

Discussion in 'Homework Help' started by J_Rod, Feb 14, 2015.

1. ### J_Rod Thread Starter Member

Nov 4, 2014
107
6

Hi all,
I just have two questions about this circuit diagram. 1. The current labelled i3 next to the 80V source - is i3 through 80V source, or through the 4kohm resistor? I can't decide if the authors meant it to be through the source or the resistor. Two of the other resistors have labelled currents, except the 1kohm resistor. Which leads to 2. Is the current through the top branch containing the 1kohm resistor 10mA, or 10mA -i3? If so, then I can determine the mesh currents.
Thanks. The problem is to find the power dissipated in the 1kohm resistor, so that's why I am trying the determine the current through it.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
You dont need the author to label the currents for you; you can do it yourself, and then you know what you mean when you define them. I see four meshes, but one of them is unambiguously predetermined for you...

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3. ### J_Rod Thread Starter Member

Nov 4, 2014
107
6
True... which one is unambiguous? The leftmost loop has current i1 +i2? The top loop has current 10mA plus whatever comes from the 80V source.

4. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
The current i3 is almost certainly in the 80V source, but unless you are specifically asked to solve for i3, then what MikeML said applies. You define the currents as you see fit.

The current in the 1kΩ resistor can not be written just in terms of the 10mA source and i3 -- you must take into account the current in the 4kΩ resistor, too. It can, however, be written in terms of the 10mA source and i1 and i2. Do you see how?

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5. ### J_Rod Thread Starter Member

Nov 4, 2014
107
6
Oh yeah, that node will draw some current into the 4kohm resistor. I would use the conservation of matter to know that sum of current entering the node equals sum of current leaving, so current through the 1kohm resistor = 10mA +i2 -i1. So I solved that using nodal analysis and used V^2 /R to find the power in the 1kohm resistor.
Thanks guys!

Last edited: Feb 15, 2015
6. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
No problem. Glad you got it.