Find currents in every loop of the circuit

[builder_user]

Homework Statement

Find currents in every loop of the circuit, find voltage on every element,build vector schelude for currents in the circuit.

Homework Equations

The Attempt at a Solution

R1=35 Ohm
R2=40 Ohm
R5=30 Ohm
R6=20 Ohm
L4=25millihenry
C4=40microfarad
E3=15*sin(300t-60)
E4=29
j6=1.25*sin(300t+75)
Z4=(RL4+RC4+r4)
I've found resuts.But when I wanted to check it with power balance the values were completely different.(Sum(e*I) !=Sum(I^2*r) at all.completely different values)

 

[gneill]

What is r4?
What are RL4 and RC4?

Is Z4 supposed to be a compound component consisting of an R and L and a C in series?

Are the other Z's just assigned the corresponding components, such as Z1 = R1, Z5 = R5, and so on?

 

[builder_user]

What is r4?
What are RL4 and RC4?

RL4 - resistance on L4
RC4 - resistance on C4
r4 - resistance on R4

Is Z4 supposed to be a compound component consisting of an R and L and a C in series?

Yes.It's just a symbol.It's easier to use this symbol in equtations.[/QUOTE]

Are the other Z's just assigned the corresponding components, such as Z1 = R1, Z5 = R5, and so on?

yes.

 

[gneill]

What value does r4 have?

 

[builder_user]

r4=25

 

[gneill]

I take it that E4 is a DC source, not an AC source, correct?

 

[builder_user]

I take it that E4 is a DC source, not an AC source, correct?

Yes,

 

[builder_user]

If current

I2=-0.278-0.227*j

I need to use I2 or 0.278+0.227*j in this equatation

I2-I3?

And I everywhere need to use module of I2 and etc.?

 

[gneill]

I don't know how your currents are assigned. Are these loop currents (how are the loops numbered?) or component currents?

 

[builder_user]

I don't know how your currents are assigned. Are these loop currents (how are the loops numbered?) or component currents?

I1- resistor r1
I2 - resistor r6
I3 - resistor r4
I2-I3 - resistor r5

 

[gneill]

So it looks like loop currents. The sign of the current reflects the direction of the current with respect to your initial assumption for loop direction. If you initially assumed clockwise flow for I2, then a negative value indicates that the current is "really" flowing counterclockwise (against the assumed direction).

So, if you're looking for the current through R5, first you need to decide which direction you want to assume the current to be flowing. In this case you'd probably like the current to be flowing "down" through the resistor towards ground. So choose the loop current that you assumed goes in the same direction and subtract the loop current that goes in the opposite direction.

 

[builder_user]

So If I change I2' sign and I2-I3=Ir5 so Ir5' direction is direction of I2?

 

[gneill]

The problem is, I don't know what directions you initially assumed for your current directions. So it's difficult for me to say. Here's a version of your circuit redrawn with the current supply replaced by its Thevenin equivalent and some assumed loop currents. How does it compare with what you've assumed?

 

[builder_user]

It's the same

 

[gneill]

Then the current flowing downward through Z5 will be I2 - I3. Leave the signs of the currents I2 and I3 alone, but look at the result of I2 - I3. If it's positive then then current "really is" flowing downwards. If it's negative, then the current actually flows upwards.

Note that these will be AC currents so knowing the 'direction' in "real life" is not particularly useful, but it is necessary to know that the direction is important to the calculations as it affects the relative phase between currents.

 

[builder_user]

so I don't need to change signs anywhere?even when I'll find voltage?

 

[gneill]

so I don't need to change signs anywhere?even when I'll find voltage?

You need not change signs. The polarity of the voltage on the component will follow from the current direction (voltage "drops" in the direction of current). If your calculated current is negative, that means it "really" flows in the opposite direction from your assumption. In such a case, the voltage drop that you calculate will be negative too (so "really" a rise rather than a drop). You can indicate this on the schematic with a voltage polarity arrow if you wish.

If it's just the magnitude of the voltage that you're interested in then you'll be taking the magnitude of the calculated values, which is always positive. Imagine that you were measuring the voltage across a component with a voltmeter. If its an AC voltage, the value you get will be the same regardless of which way you place the meter leads; you just get the magnitude of the voltage.

You could write in the voltages in magnitude + angle form if you want. It really depends upon what your instructor wants to see.

Just make sure that the relative current and voltage phases are maintained for further calculations -- such as when you want to find the power dissipated by a component.

 

[builder_user]

I'm trying to write power balance.
But Mathcad don't want to assume roots.So I can't check my results.What's the problem?

and it's seems that results are not correct. I don't understand.




Here all my calculations

 

[gneill]

I'm trying to write power balance.
But Mathcad don't want to assume roots.So I can't check my results.What's the problem?

Sorry, I don't understand. Is this a Mathcad idiosyncrasy that's causing problems?

 

[builder_user]

Sorry, I don't understand. Is this a Mathcad idiosyncrasy that's causing problems?

I've loaded my mathcad project

 

[builder_user]

I remake some calculations.Results have different signs.But they are pretty similar

 

[gneill]

We seem to have chosen a different numbering for the loops (your loop 2 is on the right). But that's a small thing.

I see in your calculation for loop 1 that you've got I3 flowing through R1. Is that true?

I've attached my version of the calculations (.mcd file in zip format) so that you can compare.

EDIT: I've replaced the attachment with a more complete, and hopefully bug-free version.

 

[builder_user]

Why you didn't use E4?

 

[gneill]

Why you didn't use E4?

If you mean, why didn't E4 appear in the AC analysis equations and power calculations, then it is because E4 is a DC source.

The extent of E4's effect on the circuit is to place a 29V bias charge on the capacitor when the circuit is first "turned on". After that the capacitor blocks further DC current from flowing.
Also, being an ideal voltage source it acts as a short circuit for AC current -- it just "disappears" as far as AC analysis of the circuit is concerned.

 

[builder_user]

and one more thing.
why sum of all powers on every E has sign 1 and sum of all powers on evey Z has sing -1?

They must have different signs?I know that for DC they should be equal

 

[gneill]

and one more thing.
why sum of all powers on every E has sign 1 and sum of all powers on evey Z has sing -1?

They must have different signs?I know that for DC they should be equal

It is a convention based upon the definition of the power consumed by a device. If you look at a general two-terminal device, the power consumed is defined as the product of the current flowing into the device and the voltage across the device: P = I*V. But what happens when the two-terminal device happens to be a voltage supply with current coming *out* of the + terminal? Then the devices' current is negative, and the product I*V is also negative, showing that the supply is "losing" power to the circuit -- that is, it's supplying energy to the circuit rather than consuming it.

Things that consume power (like resistances) are assigned a positive value. Things that are putting power into the circuit (like most sources -- although it's certainly possible for a source to consume power! So beware) will have negative values. That way when you sum up all the sinks and sources, the result should be zero for energy balance.