Working out freq for circuit to be in phase

In summary, the current and voltage in this AC circuit are in phase when the frequency is equal to the inductance and capacitance values.
  • #1
craka
20
0

Homework Statement



AC circuit Vs is 100V 50Hz and is used as reference.
there are two impedences in parallel, made up of a 5ohm resistor and an inductor reactance = +j2 ohm that is parallel with 3 ohm resistor and a inductor reactance = -j3 ohm (though I though negative imaginary part was for capactance?)

voltage across the 3 ohm resistor is 45 Volts

The branch with the resistor and inductor is called I1 , branch with resistor and capacitor is I2
Is is the total current from supply before the parallel branch.

Please see attachment of circuit

Homework Equations



I2 = v/I angle = arctan x/r

Is = Vs/Zt = Vs * 1/ (Z1 parallel Z2) = Z1Z2/(Z1+Z2)

The Attempt at a Solution



I2 = 45/3 =15A angle = artan -3/3 = -45 degrees

magnitude of Vs = 15 * sqrt(3^2+3^2) = 63.64 volts

Z1 = 5+2j = 5.385 angle 21.8 deg Z2 = 3-3j = 4.234 angle -45 deg

Z1//Z2 = {(5.385 angle 21.8) * ( 4.234 angle -45)}/ {(5+2j)+(3-3j)}

Z1//Z2 = 22.849 angle -23.2 deg / 8.06 angle -7.125

Z1//Z2 = 2.835 angle -16.075 deg

Is = 63.64/(2.835 angle -16.075) = 22.45 angle -16.08 degFrom here is where I'm lost on how to calculate the frequency needed to get this circuit to have current and voltage in phase (zero degrees)
 

Attachments

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  • #2
For voltage and current to be in phase, the impedance must be a pure resistance, so the imaginary part of your impedance must be zero.
Since you know the impedances of capacitor and inductor at 50Hz, you can calculate their values.
Write Z1, Z2 and Z1//Z2 as functions of the known values of resistances, inductances and capacitances and the unknown value of frequency. Then calculate the frequency for which the imaginary part of the impedance is zero.
 
  • #3
I had worked out the inductance values of components for the known frequency of 50Hz howerver I'm unsure what you mean by "write Z1, Z2 and Z1//Z2 as functions of the known values of resistance, and inductance capacitance and the unknonwn value of frequency".
 
  • #4
craka said:
I had worked out the inductance values of components for the known frequency of 50Hz howerver I'm unsure what you mean by "write Z1, Z2 and Z1//Z2 as functions of the known values of resistance, and inductance capacitance and the unknonwn value of frequency".

If you call f the frequency, [tex]Z_1=R_1 + 2\pi f L_1[/tex].
You have a similar expression for [tex]Z_2[/tex] and for the parallel of the two. The only unknown is f, that you calculate by making the imaginary part of the impedance equal to zero.
 
  • #5
So I have Z1 = 5 + 2pifL and Z2 = 3 + 2pifl but what is the expression for Z1//Z2 with using the above value of Z1 and Z2 ??
 
  • #6
craka said:
So I have Z1 = 5 + 2pifL and Z2 = 3 + 2pifl but what is the expression for Z1//Z2 with using the above value of Z1 and Z2 ??

No,
[tex]Z_1 = 5 +j2\pi f L[/tex]
[tex]Z_2 = 3+\frac{1}{j2\pi f C}[/tex]

and
[tex]Z_1 // Z_2 = \frac{Z_1 Z_2}{Z_1 + Z_2}[/tex]
 
  • #7
from that i get this

[tex]
\begin{array}{l}
z_1 //z_2 = \frac{{(5 + j2\pi fL) \times (3 + j\frac{1}{{2\pi fC}})}}{{(5 + j2\pi fL) + (3 + j\frac{1}{{2\pi fC}})}} \\
= \frac{{15 + j\frac{5}{{2\pi fC}} + j6\pi fL + \frac{L}{C}}}{{8 + j2\pi fL + j\frac{1}{{2\pi fC}}}} \\
\end{array}
[/tex]
Again I'm stuck. Could you please give me some more assistance thankyou.
 
Last edited:
  • #8
craka said:
from that i get this

[tex]
\begin{array}{l}
z_1 //z_2 = \frac{{(5 + j2\pi fL) \times (3 + j\frac{1}{{2\pi fC}})}}{{(5 + j2\pi fL) + (3 + j\frac{1}{{2\pi fC}})}} \\
= \frac{{15 + j\frac{5}{{2\pi fC}} + j6\pi fL + \frac{L}{C}}}{{8 + j2\pi fL + j\frac{1}{{2\pi fC}}}} \\
\end{array}
[/tex]
Again I'm stuck. Could you please give me some more assistance thankyou.

Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. In this way only the numerator will be complex.
Equate the imaginary part of the numerator to zero and calculate the value of f.
 

1. How do I determine the frequency required for a circuit to be in phase?

The frequency required for a circuit to be in phase is determined by the circuit's impedance and the frequency response of the components within the circuit. It can be calculated using the formula f = 1 / (2πRC), where R is the resistance and C is the capacitance of the circuit. This formula assumes a simple RC circuit, but more complex circuits may require more advanced calculations.

2. What is the significance of having a circuit in phase?

Having a circuit in phase means that the voltage and current within the circuit are in sync and reach their maximum and minimum values at the same time. This is important for efficient power transfer and reduces losses in the circuit.

3. How does changing the frequency affect the phase of a circuit?

Changing the frequency of a circuit can shift the phase angle, which is the difference in phase between the voltage and current. As the frequency increases, the phase angle decreases and the circuit becomes more in phase. Similarly, decreasing the frequency increases the phase angle and the circuit becomes more out of phase.

4. Can a circuit be in phase at any frequency?

No, a circuit can only be in phase at a specific frequency. This frequency is known as the resonant frequency and is determined by the circuit's components. At this frequency, the circuit has the lowest impedance and is most efficient.

5. How can I adjust the frequency of a circuit to achieve phase alignment?

The frequency of a circuit can be adjusted by changing the values of the components within the circuit. For example, altering the resistance or capacitance can change the resonant frequency and bring the circuit into phase. Additionally, using a frequency generator or oscillator can provide a variable frequency signal to test and adjust the phase of a circuit.

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