Designing Inductor Value for Coupled RLC Circuit

Click For Summary
Designing an inductor value for a coupled RLC circuit requires considering input and output voltages, output current, and coupling coefficients. While no single equation applies universally, methods such as using resonance frequency and voltage transfer functions can help determine the appropriate inductance. The resonance frequency can be calculated with f = 1/(2π√(LC)), allowing for the calculation of L when capacitance is known. Additionally, the quality factor (Q) of the circuit, which influences resonance sharpness, can be calculated using Q = ωL/R. Overall, a combination of these approaches will aid in accurately designing the inductor for the circuit.
abhierao
Messages
7
Reaction score
0
hello,

i am deriving an equation for designing an inductor in a coupled RLC circuit. is there an equation for calculating inductor value by knowing the input and output voltages & output current of the RLC circuit ?
i have tried KVL and KCL equations but in vain.
please help

thank you
 
Engineering news on Phys.org
Using KVL and KCL you end up with a 2nd Order Differential Equation. This equation has to be solved in order to find the time behavior of the system.

Have you learned about Laplace Transforms yet? Or do you know how to solve differential equations?
 
i am running a steady state solution so i do not have to worry about transient solution & differential equations. but if you suggest me to run through differential equation then that's not a problem, yes i do know laplace transforms & i have solved differential equation for current in a coupled rlc circuit
 
http://www.ee.bgu.ac.il/~intrlab/lab_number_7/Two inductively coupled RLC circuits.pdf
this is one link that i can link to my work. the circuit is similar but with a load resistance at output where i am measuring my voltage. he has some equations by assumptions but i have made use of KVL to get without those assumptions.
i know input voltage 20volts, output power 15v , 1A or 10v 1.5A. k i can choose anywhere close to 1 or lesser values, i want to design inductor with these values. i have frequency of 100 hz.
please help
 
for your question. Designing an inductor value for a coupled RLC circuit can be a complex task, as it involves considering various factors such as the input and output voltages, output current, and the coupling coefficient between the inductor and the rest of the circuit. While there is no single equation that can be used to calculate the inductor value in all cases, there are a few approaches that can be taken to determine the appropriate value for your specific circuit.

One method is to use the resonance frequency of the circuit, which is the frequency at which the inductive and capacitive reactances are equal. This can be calculated using the equation f = 1/(2π√(LC)), where L is the inductance and C is the capacitance. By knowing the desired resonance frequency and the capacitance value, you can solve for the required inductance value.

Another approach is to use the voltage transfer function of the circuit, which relates the input and output voltages. By considering the desired output voltage and the input voltage, you can calculate the inductance value using the equation L = Vout/(ωIout), where ω is the angular frequency and Iout is the output current.

It is also important to consider the quality factor (Q) of the circuit, which is a measure of the circuit's ability to store and release energy. A higher Q value typically results in a sharper resonance peak and better selectivity. The Q value can be calculated using the equation Q = ωL/R, where R is the resistance in the circuit.

In summary, there is no one equation that can be used to calculate the inductor value for a coupled RLC circuit, but by considering factors such as the resonance frequency, voltage transfer function, and quality factor, you can determine the appropriate value for your specific circuit. I hope this helps in your design process. Best of luck!
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
44
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K