Equivalent resistance and inductance

In summary, U is meant to represent an AC or DC current source, but if it's a DC source then things will be more difficult.
  • #1
Trainee28
9
0

Homework Statement


Hello, I am going to ask a rather simple question yet I wasn't able to solve. I have a circuit which I wanted to find out about the equivalent resistance and inductance. I know how to find the equivalent resistance of two resistors in parallel as well as two inductors in parallel, but when it is one resistor in series with an inductor and the two components in parallel with another similar branch, I couldn't figure out the calculation. I wanted to find out the time constant of the circuit actually. If it is one resistor in series with an inductor, it will be L/R.

Can anyone help me please?
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Homework Equations

The Attempt at a Solution


Zeq=Req+Leq=[(R1*R2)/(R1+R2)]+[(L1*L2)/(L1+L2)]
 
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  • #2
You'll get two separate time constants in general. Adding the branches like this does not work.
 
  • #3
If you haven't covered phasors you don't want to try to solve this.
 
  • #4
Hi Trainee28, is U meant to represent an AC or DC current source? If it's an AC source with a known frequency then you could use a complex impedance approach, which is pretty straightforward. If it's a DC source then things will be more difficult, as the others have mentioned.
 
  • #5
Even the left side of the shown equation is wrong. You must not add R and L. Instead, for L you must use the impedance jwL.
Hence, Z1=R1+jwL1 and Z=Z1||Z2.
As mentioned before, it is a - somewhat involved - however, straightforward calculation. At the end you have to find Re(Z) and Im(Z).
 
  • #6
Assuming the big arrow has no special meaning, you have two independent circuits with the same constant voltage (AC or DC) in both. That is easy to solve, but you cannot add the resistances and inductances. Currents can be added.
 
  • #7
gneill said:
Hi Trainee28, is U meant to represent an AC or DC current source? If it's an AC source with a known frequency then you could use a complex impedance approach, which is pretty straightforward. If it's a DC source then things will be more difficult, as the others have mentioned.

It is a 20kHz PWM voltage signal of 12V, with a duty cycle of 50%. So that will make it an average voltage of 6V. But I don't know how to relate a PWM voltage source signal to a complex impedance calculation.
 
  • #8
mfb said:
Assuming the big arrow has no special meaning, you have two independent circuits with the same constant voltage (AC or DC) in both. That is easy to solve, but you cannot add the resistances and inductances. Currents can be added.
Actually I am trying to simulate a voltage chopper. With U(t) a 20kHz PWM voltage source 12V and 50% duty cycle, I know that the current I(t) will rise during the conduction part of the voltage signal and fall when the voltage drop to 0. With only one branch of resistance in series with an inductor, I know that the current I(t) will rise in exponential form towards the value Uaverage/R(with Uaverage=12V*50%), and oscillate around this value in steady state due to the alternative high and low state of voltage U.

I can therefore find out the rise time of current to 65% of its maximum value and deduce the inductance value because rise time=L/R. But when comes to two branches of (R+L)//(R+L), I have no idea how the circuit will response.
 
  • #9
Any sort of square or rectangular waveform comprises many harmonics (Fourier), so there's no single frequency to use for purposes of analysis. You'll have to turn to differential equations or Laplace transform methods then.

There are some things you can do to simplify the analysis if the time constants associated with the individual branches are small enough so that any transient behavior dies out between transitions of the source voltage. Since the branches parallel the voltage source they can be treated separately at the initial turn-on. After that you need to consider carrying over any existing circuit conditions (currents) at the following transitions as new initial conditions. You may need to follow the circuit over several cycles in order to reach its steady operating behavior where the initial conditions of each cycle are the same.

As a hint, consider that a zero-voltage voltage source is indistinguishable from a short circuit. Redraw the circuit accordingly for those periods of time.

Personally, if this were not a homework question I'd just turn to a simulation package like LTSpice and be done with it in a few minutes of effort!
 

Related to Equivalent resistance and inductance

1. What is equivalent resistance and inductance?

Equivalent resistance and inductance are electrical properties that describe the behavior of a circuit. Equivalent resistance (Req) refers to the total resistance a circuit experiences when multiple resistors are connected in a series or parallel configuration. Equivalent inductance (Leq) refers to the total inductance a circuit experiences when multiple inductors are connected in series or parallel.

2. How is equivalent resistance calculated in a series circuit?

In a series circuit, the equivalent resistance is calculated by simply adding the individual resistances together. This is because the current has no choice but to flow through each resistor in succession, resulting in a cumulative effect.

3. How is equivalent resistance calculated in a parallel circuit?

In a parallel circuit, the equivalent resistance is calculated using the following formula: 1/Req = 1/R1 + 1/R2 + ... + 1/Rn. This is because the current has the option to flow through multiple paths, resulting in a decrease in total resistance.

4. How is equivalent inductance calculated in a series circuit?

In a series circuit, the equivalent inductance is calculated by simply adding the individual inductances together. This is because the magnetic field created by each inductor adds together, resulting in a cumulative effect.

5. How is equivalent inductance calculated in a parallel circuit?

In a parallel circuit, the equivalent inductance is calculated using the following formula: 1/Leq = 1/L1 + 1/L2 + ... + 1/Ln. This is because the magnetic fields created by each inductor add together, resulting in a decrease in total inductance.

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