Finding Frequency of 20mH Inductor & 30 Ohm Resistor in Parallel

In summary, To find the frequency omega in rad/s, we use the impedance formula for a parallel circuit and set it equal to the given phase shift in degrees. However, this approach is incorrect as we need to convert the phase shift to radians and rearrange the impedance equation to solve for omega. Additionally, we need to make the denominator of the impedance equation real before proceeding with the calculation.
  • #1
mugzieee
77
0
im given a a 20-mH inductor and a 30 ohm resistor in parallel. Z_in is 25 degrees. and I am asked to find the frequency omega in rad/s here's what i try to do:
25=(jw.02*30)/(jw.02+30), and solve for w, but i don't get the right answer.
what is it that I am doing wrong?
 
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  • #2
mugzieee said:
im given a a 20-mH inductor and a 30 ohm resistor in parallel. Z_in is 25 degrees. and I am asked to find the frequency omega in rad/s here's what i try to do:
25=(jw.02*30)/(jw.02+30), and solve for w, but i don't get the right answer.
what is it that I am doing wrong?

You're given the phase shift (in degrees). Remember that the phase shift is the argument of the complex impedance of the combination.

So you started out determining the impedance correctly :

[tex]Z = Z_L // R = \frac{j\omega L}{R + j\omega L}[/tex]

but you then equated that to 25 degrees, which makes no sense. Keep in mind that Z is a full complex number with a magnitude and an argument. You should rearrange Z to the form [tex]Z = re^{j\theta}[/tex] where [tex]\theta[/tex] is the radian equivalent of 25 degrees (25/180*pi) Find an expression for the argument in terms of the arctangent of a ratio between the resistance and the inductance times omega. That's the equation you need to solve for omega.

The first thing you should do in that expression for Z is to make the denominator real.
 
  • #3


It seems like you are trying to use the impedance formula for a parallel circuit, which is not applicable in this case. In a parallel circuit, the impedance of the components is calculated using the formula Z = (1/Z1 + 1/Z2 + ... + 1/Zn)^-1, where Z1, Z2, etc. are the impedances of the individual components. However, in this case, we are given only the inductance and resistance values, and the formula for impedance in a series circuit is Z = sqrt(R^2 + (wL)^2). Therefore, to find the frequency omega, we need to rearrange the formula as w = sqrt(Z^2 - R^2)/L. Plugging in the given values, we get w = sqrt((25)^2 - (30)^2)/20mH = 25 rad/s.
 

1. How do you calculate the total impedance of a parallel circuit with a 20mH inductor and a 30 Ohm resistor?

The total impedance of a parallel circuit with a 20mH inductor and a 30 Ohm resistor can be calculated using the formula ZT = 1 / (1/ZL + 1/ZR), where ZL is the impedance of the inductor and ZR is the impedance of the resistor. In this case, ZL = jωL = j(2πfL) = j(2π*50*0.02) = j6.28Ω and ZR = R = 30Ω. Plugging these values into the formula, we get ZT = 1 / (1/j6.28 + 1/30) = 29.94Ω.

2. What is the angular frequency of a 20mH inductor?

The angular frequency of a 20mH inductor can be calculated using the formula ω = 1/√(LC), where L is the inductance and C is the capacitance. In this case, L = 20mH = 0.02H and C = 1/ω2 = (1/(2πf))2 = (1/(2π*50))2 = 0.000064H. Plugging these values into the formula, we get ω = 1/√(0.02*0.000064) = 50 rad/s.

3. How does the impedance of an inductor and a resistor differ in a parallel circuit?

In a parallel circuit, the impedance of an inductor and a resistor differ in the way they combine. The impedance of a resistor remains constant regardless of the frequency, while the impedance of an inductor increases with frequency. This means that at low frequencies, the inductor will have less impedance compared to the resistor, but at high frequencies, the inductor will have more impedance.

4. How do you calculate the current through a parallel circuit with a 20mH inductor and a 30 Ohm resistor?

The current through a parallel circuit can be calculated using the formula I = V/ZT, where V is the voltage and ZT is the total impedance of the circuit. In this case, we can use the value of ZT calculated in question 1 and the voltage can be given. For example, if the voltage is 10V, then the current would be I = 10/29.94 = 0.334A.

5. How does the phase difference between the current and voltage change in a parallel circuit with an inductor and a resistor?

In a parallel circuit, the phase difference between the current and voltage changes with frequency. At low frequencies, the current and voltage are in phase, meaning they have the same phase angle. However, as the frequency increases, the current leads the voltage, meaning it has a higher phase angle compared to the voltage. This is because the impedance of the inductor increases with frequency, causing the current to be affected more by the inductor than the resistor.

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