Currents in electrical circuit

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KamilAussie

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Czech Republic
Hi, I'm struggling to solve my homework again. The task is to find out values of all three currents and the source power.

The circuit looks like this:

obvod.jpg
I was given impedance of each component and information about AC source (voltage 66 V, angle 1.09 rad, sinus characteristic).

My solution:
Few components' impedance is 0Ω, that's why they aren't in my diagram.

1) Calculating impedance of whole circuit from the source. (Paralel combination of i3 (red) and i7 (orange) part + i1 (green) part in a series)

2) Calculating I (I=U/Z), which is the same value as I1?

3) Calculating I7 from a loop (II. Kirchhoff's law)

4) Calculating I3 from point C (I. Kirchoff's law)

I entered the values into a school system and all of them are wrong. I guess I can't calculate I1 the way I did?

I'm trying to upload my solution but it keeps failing. It's gonna be in section below.
Thanks for helping me.

Okay, my solution is here -> https://s18.postimg.org/8lnx1q3bt/20161125_172949.jpg
 
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Ingenieur

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Location
Earth
I1 looks correct
rather than solving for I7 using voltage solve for I3 and I7 using a current divider

I3 = I1 x Z7/(Z3+Z7)

I7 = I1 x Z3/(Z3+Z7)

and only work to 4 decimal places for numbers and 1 for angles
 
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KamilAussie

Member
Location
Czech Republic
I know it's easier using current divider, I did it first time like that but when I got incorrect results I wanted to try another method. The values are the same tho.

I did the math again with only 4 decimal for numbers and 1 decimal for angles. That damn system says it's incorrect anyway! :happysad:


Another question.. When solving real power P=U*I*cos(angle) and reactive power Q=U*I*sin(angle), is it correct to use only size of voltage and current without angle?
 
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Ingenieur

Senior Member
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here's what I get (rounded):
I1 = 0.32/94 deg = -0.023 +0.32j
I3 = 0.21/-18 deg = 0.196 - 0.065j
I7 = 0.44/120 deg = -0.22 + 0.38J

S = 21.1/-28 deg = 18.6 - 9.9j
so
P = 18.6 W
Q = 9.9 var (capacitive)

close to yours with the exception of power(s)
 
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GoldDigger

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Retired PV System Designer
KamilAussie said:
I know it's easier using current divider, I did it first time like that but when I got incorrect results I wanted to try another method. The values are the same tho.

I did the math again with only 4 decimal for numbers and 1 decimal for angles. That damn system says it's incorrect anyway! :happysad:


Another question.. When solving real power P=U*I*cos(angle) and reactive power Q=U*I*sin(angle), is it correct to use only size of voltage and current without angle?
Click to expand...

Just recognize that the (angle) present in that formula is not the angle specfied in the statement of the problem:
I was given impedance of each component and information about AC source (voltage 66 V, angle 1.09 rad, sinus characteristic).
Click to expand...

Instead it is the difference in phase angle between the given voltage and the current calculated from your network solution.
 
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