What is the Current at Node Y in a Circuit with Kirchhoff's Current Law?

[Paymemoney]

Problem i am stuck on:
In 10 seconds, an observer - Willy Nilly - notices that 35 coulombs of charge leaves node "n" in this circuit, heading for node "x". (Vn is the voltage at node "n", etc.) In the same ten seconds, 22 coulombs of charge leaves node "n" heading for node "z". Determine the current, Iy.

http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Basic/Circuit1B1.gif

I have said that
Ix=35
Iz=22
Iy=Ix-Iz

so Iy=13

P.S

 

[vk6kro]

The current into a junction must equal the current flowing out of it.

Does your solution show this?

 

[Paymemoney]

the problem is i don't have solutions.

 

[snarfherder]

the problem is i don't have solutions.

Current entering node n = current leaving node n? sum the currents leaving this node and you'll get the current entering this node. Right now you have a net loss of current from a single node.

think of it as a zero sum equation:

0 = (Inx + Iny + Inz)
0 = (-35 + x + -22)

 

[Paymemoney]

so
0=-57+x
x=57

 

[willem2]

so
0=-57+x
x=57

Now you need to get the signs right.
If charge moves in the direction of the arrows then you have a positive current.
Charge leaving node n for node x flows in the same direction as the arrow next
to I_x, so I_x is positive. Same for I_y.

Also current is in Ampere which is coulomb/second so 35 coulombs in 10 seconds is only 3.5 A

 

[snarfherder]

Now you need to get the signs right.
If charge moves in the direction of the arrows then you have a positive current.
Charge leaving node n for node x flows in the same direction as the arrow next
to I_x, so I_x is positive. Same for I_y.

sign is very important, but in this case charge moving is comprised of electrons, and electrical current is defined to be opposite their flow. 35 [C] flowing from n to x over 10 seconds means that a current of 3.5 A flows FROM x TO n. Similarly, 57 [C] from y to n means 5.7 [A] FROM n TO y. Iy is therefore negative, Ix is positive, Iz is positive, but I personally find it easiest to think first in terms of electrons moving and then to realize that current flow is defined (originally by ben franklin) to be opposite that of electrons

Also current is in Ampere which is coulomb/second so 35 coulombs in 10 seconds is only 3.5 A

yes current = Ampere = [C/s], assuming you're not supposed to sum current over the 10 second time period

 

[Paymemoney]

ok thanks guys for answering the question i understand now.