Engineering How to find current and current consumption of entire circuit

[aruwin]

I need help in answering these questions.This is not homework, this is exercise.

Given that current i1(t)= (√2)*sin2t[A]

1. Find the current i2(t).
2.Find the effective current(current consumption) in the entire circuit

 

[vk6kro]

Have you tried to do it yourself?

Why not show your working and maybe we can see what you don't understand about it?

It doesn't matter that it isn't homework. If it is like homework, then the homework rules apply.

 

[jono20]

Well at first glance of the values I'd guess that all of the currents are going to be somewhere between very high and holy-cow!



AC RLC circuits are solved similarly to resistive DC circuits.

Find the impedance of each component, and do typical series and parallel additions.

The trick is, since it's AC and inductive current leads and capacitive current lags, you'll need to do those additions in the complex plane.

 

[vk6kro]

Maybe not as high as you think.

Can you say what the frequency and peak amplitude of the current i1 is?

AC and inductive current leads and capacitive current lags.

Better check this one...

 

[aruwin]

I got this for number 1. Please correct my mistakes.Please.

 

[reddvoid]

I got like this. . .please correct me if wrong

 

[vk6kro]

I1 is given so you don't have to derive it.

The reactance of a 0.5 H inductor at a frequency of 0.318 Hz is close to 1 ohm, not 4 ohms.


Also,
I don't think you can divide currents like that.

Suppose you put a 5 ohm and a 10 ohm resistor in parallel and the total current was 10 amps.
The 10 ohm resistor would be carrying 3.33 amps and the 5 ohm resistor would be carrying 6.66 amps.

So could you say that the current in the 10 ohm resistor was 10 amps times 5/10 ? No, that would be 5 amps.

 

[reddvoid]

I1 is given so you don't have to derive it.

The reactance of a 0.5 H inductor at a frequency of 0.318 Hz is close to 1 ohm, not 4 ohms.


Also,
I don't think you can divide currents like that.

Suppose you put a 5 ohm and a 10 ohm resistor in parallel and the total current was 10 amps.
The 10 ohm resistor would be carrying 3.33 amps and the 5 ohm resistor would be carrying 6.66 amps.

So could you say that the current in the 10 ohm resistor was 10 amps times 5/10 ? No, that would be 5 amps.

sorry that was a real blunder

 

[NascentOxygen]

That looks better.

Are there two students from the same class posting in this thread?

You were given the current as a sinusoid; I'm wondering why you changed it to a cosinusoid?

 

[gneill]

How did the 4j impedance of the upper inductor turn into 3j in the equation for I1?

 

[vk6kro]

The answer is correct but the method is still wrong.

Going back to the example above with the 10 ohm and 5 ohm resistors with a total current of 10 Amps.

The parallel combination of 5 and 10 ohms is 3.333 ohms.

The current in each resistor is given by 10 amps times 3.333 / R.
So, if R is 10 ohms, the current through it is 10 * 3.333 / 10 ...or 3.33 amps.
If R is 5 ohms, the current is 10 * 3.333 / 5 ...or 6.6666 amps.

So, you don't add the impedances in the loop. You need to get the parallel combination of them.

An easier way would be to derive the current in the 0.5 F capacitor and just add the two currents when you have both currents.

 

[reddvoid]

The answer is correct but the method is still wrong.

Going back to the example above with the 10 ohm and 5 ohm resistors with a total current of 10 Amps.

The parallel combination of 5 and 10 ohms is 3.333 ohms.

The current in each resistor is given by 10 amps times 3.333 / R.
So, if R is 10 ohms, the current through it is 10 * 3.333 / 10 ...or 3.33 amps.
If R is 5 ohms, the current is 10 * 3.333 / 5 ...or 6.6666 amps.

So, you don't add the impedances in the loop. You need to get the parallel combination of them.

An easier way would be to derive the current in the 0.5 F capacitor and just add the two currents when you have both currents.

but we can do like this
current through 10 ohms is 10A*[5/(5+10)] and
current through 5 ohms is 10A*[10/(5+10)]
/* current division principle*/

 

[gneill]

 

[vk6kro]

but we can do like this
current through 10 ohms is 10A*[5/(5+10)] and
current through 5 ohms is 10A*[10/(5+10)]
/* current division principle*/

Yes, you are quite right.

That is a much easier way of doing it.

 

[reddvoid]

You were given the current as a sinusoid; I'm wondering why you changed it to a cosinusoid?

I think using cosinusoidal while working with phasors makes it easier to work .
but here if we use sinusoidal final answer we get as 4.24i
that is 4.42arg(90)
why is that ?