# Calculate the power of the coils in this RLC circuit

• Engineering
• Edy56
In summary, the power dissipated in the coils is imaginary because the coils have no real series resistance.
Edy56
Homework Statement
I need to solve for power of the coils
Relevant Equations
none

So my attempt was
I1*jXl1+I2*jXm=I2*(jXl2-jXc2) + I1*jXm
because they are parallel so they should have the same voltage.
I got
I1=-3*I2.
I know that J=6-j4 and that J=I1+I2 so I just plugged in what I got and I ended up getting the result for I2 and I1.
But they are not correct.
Why?
I am given R and I never use it so I assume it has something do do with it.

Please, take a moment to learn LaTeX Guide. And use it to re-write your equations.

DaveE and berkeman
How can there be any power dissipated in the coils?
They have no real series resistance, so the product of v, and quadrature i, is imaginary = VAR.

Baluncore said:
How can there be any power dissipated in the coils?
They have no real series resistance, so the product of v, and quadrature i, is imaginary = VAR.
In the circuit determine the complex powers of the coupled coils.

If Xm=0 then I1*jXL1=I2*j(XL2-XC).

Where is Xm located? It could be in Слика 1?

If Xm=0 then I1*jXL1=I2*j(XL2-XC).

Where is Xm located? It could be in Слика 1?
no Xm= X12 which is mutual coupling.
to my knowledge, since this is coupled coils
Ul1=jXl1*I1 +\- jXm*I2
Ul2=jXl2*I1 +\- jXm*I1

Something like this?

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Something like this?
I am not too sure because we havent been taught that we can do that, but going from the videos I have seen, it would do the same thing

Are you sure they are coupled inductors? It seems odd that they wouldn't tell you the turns ratios or coupling coefficients.

This is the diagram of a transformer. Then a more simplified diagram can be used:

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Then, if it is the diagram of a transformer, the left part is the primary and the right part is the secondary.
The diagram is viewed from primary side. That means, the secondary actually circuit elements [XL2 and XC] are multiplied by the square of number of turns of primary divided by number of turns of secondary.
The current through the primary and secondary windings are inversely proportional with the number of turns in windings.
N1*I1+N2*I2=N1*Io
Io is the primary current in absence of secondary current.
If N1=N2 [ since the reference to primary is already done] I1+I2=Io
Neglecting Io [usually it is about 1-3% I1] I1=-I2.
So, both current are I1=J and I2=-J
The reactances shown above: XL1 and XL2 , are connected with the leakage of magnetic flux from each winding [primary and secondary]
As it is said above, the power from coils-only in XL1 and XL2- it is only a reactive power.
Active power is only from R [ the R on the secondary side is also neglected]

this is getting a bit too complex for my level lol

## What is the formula to calculate the power in an RLC circuit?

The power in an RLC circuit can be calculated using the formula $$P = VI \cos(\phi)$$, where $$V$$ is the voltage, $$I$$ is the current, and $$\cos(\phi)$$ is the power factor. The power factor is the cosine of the phase angle $$\phi$$ between the voltage and current.

## How do I determine the phase angle in an RLC circuit?

The phase angle $$\phi$$ in an RLC circuit can be determined using the impedance values. It is given by $$\tan(\phi) = \frac{X_L - X_C}{R}$$, where $$X_L$$ is the inductive reactance, $$X_C$$ is the capacitive reactance, and $$R$$ is the resistance. The phase angle $$\phi$$ can then be found using $$\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$$.

## What are the components of power in an RLC circuit?

In an RLC circuit, the power can be divided into three components: real power (P), reactive power (Q), and apparent power (S). Real power is the actual power consumed by the resistive components, reactive power is associated with the energy stored and released by the inductive and capacitive components, and apparent power is the product of the RMS voltage and RMS current.

## How do inductance and capacitance affect the power in an RLC circuit?

Inductance and capacitance affect the reactive power in an RLC circuit. Inductive reactance ($$X_L$$) and capacitive reactance ($$X_C$$) create a phase difference between voltage and current, which affects the power factor. A higher phase difference reduces the real power delivered to the load, while the reactive power increases.

## Can I calculate the power of just the coil (inductor) in an RLC circuit?

The power associated with just the coil (inductor) in an RLC circuit is the reactive power ($$Q_L$$), which can be calculated using $$Q_L = I^2 X_L$$, where $$I$$ is the current through the inductor and $$X_L$$ is the inductive reactance. Note that this power is not dissipated but rather oscillates between the inductor and the circuit.

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