Electrical Circuits Reactive Networks Question

In summary, at low frequencies, the inductors act as short circuits and the capacitors act as open circuits, resulting in no current flow beyond Rs and a voltage of zero at the load. At high frequencies, the inductors act as open circuits and the capacitors act as short circuits, allowing current to flow through Rs and creating a voltage divider with the load. Parts C and D involve using transfer functions and calculating the resonant frequencies, but further information is needed to solve these parts.
  • #1
johnwillbert82
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Homework Statement


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Homework Equations


At low voltage frequency (inductors) L become short circuit and (capacitors) C becomes open circuit
At high voltage frequency (inductors) L become open circuit and (capacitors) C becomes short circuit

The Attempt at a Solution



Part A
At low frequency, all the capacitors are open, so no current can't run beyond Rs. Hence current at the load is zero so voltage is also zero

Part B
At high frequency, all the inductors are open, so the only path for the current is through Rs and around all the loops at the top where all the capacitors use to be (skipping all resistors)

So now it is the voltage supply Rs and the load in series, so a voltage divider can be used to workout the voltage across the load

V1 = V(Supply) x R1 / (R1 + R2)

v(Load) = 10 x 35 / (35 + 30) = 5.384615

Part C
I am assuming the angular frequency has no effect on the voltage as T = 1/f and omega = 2PIf so the frequency cancels out
This is most likely wrong

Part D
Using the same value from Part C, I would use P = V^2/R where R is 35

I think I have the first two parts correct and the last two parts wrong
any help?
 
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  • #2
johnwillbert82 said:
Part A
At low frequency, all the capacitors are open, so no current can't run beyond Rs. Hence current at the load is zero so voltage is also zero
At low frequencies, current won't see a path through the capacitors, so why wouldn't current flow through the resistors to reach the load?
 
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  • #3
Because of the short circuits were all the inductors use to be, thus the resistors and the load are isolated
 
  • #4
At low frequencies the inductors will quickly curtail any signal that tries to take the resistor route; They'll look like shorts to ground along the path.

Parts (c) and (d) look like you're supposed to know something significant about the "unit cell" behavior, since there are three identical filter sections in cascade. I can identify them as so-called bridged T filters. Perhaps there's a property of them that allows you to use transfer function approach?
 
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  • #5
gneill said:
At low frequencies the inductors will quickly curtail any signal that tries to take the resistor route; They'll look like shorts to ground along the path.

Parts (c) and (d) look like you're supposed to know something significant about the "unit cell" behavior, since there are three identical filter sections in cascade. I can identify them as so-called bridged T filters. Perhaps there's a property of them that allows you to use transfer function approach?

So am I right to assume since the signals are "shorted to ground", that after current passes through Rs and the first resistor, it is "curtailed" by any inductors and no other resistors or the load receives any current, which is why I put zero.

In regards to C and D, don't think I've learned about any of that yet but thanks
 
  • #6
johnwillbert82 said:
So am I right to assume since the signals are "shorted to ground", that after current passes through Rs and the first resistor, it is "curtailed" by any inductors and no other resistors or the load receives any current, which is why I put zero.
Yes.
 
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  • #7
johnwillbert82 said:
So am I right to assume since the signals are "shorted to ground", that after current passes through Rs and the first resistor, it is "curtailed" by any inductors and no other resistors or the load receives any current, which is why I put zero.
That is correct, and an improvement on how you explained it earlier.

In regards to C and D, don't think I've learned about any of that yet but thanks
Any idea how you might be expected to attack this problem? Have you looked at something similar to it in class, perhaps calculated the bridged-T resonant frequencies?
 
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  • #8
NascentOxygen said:
That is correct, and an improvement on how you explained it earlier.Any idea how you might be expected to attack this problem? Have you looked at something similar to it in class, perhaps calculated the bridged-T resonant frequencies?

Alright thanks for the help and clarification :)

Nope, no idea at all on these, going to review my notes and lectures and see if there's anything I've missed
 

1. What is the difference between an electrical circuit and a reactive network?

An electrical circuit is a closed loop of conductive materials that allows the flow of electric current. A reactive network, on the other hand, is a circuit that contains reactive components such as inductors and capacitors, which store and release energy in the form of magnetic and electric fields.

2. How do inductors and capacitors affect the behavior of an electrical circuit?

Inductors and capacitors are reactive components that introduce impedance, or resistance, in an electrical circuit. This affects the flow of current and can cause phase shifts between voltage and current, resulting in the circuit's overall behavior.

3. What is the significance of resonance in reactive networks?

Resonance is the phenomenon where the reactive components in a circuit cause the voltage and current to be in phase, resulting in a large amount of energy being stored and exchanged between the inductor and capacitor. This can lead to high voltages and can be utilized in applications such as tuning circuits and filters.

4. How do you calculate the impedance of a reactive network?

The impedance of a reactive network can be calculated using the formula Z = √(R² + (XL - XC)²) where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. This calculates the total opposition to current flow in the circuit.

5. What are some common applications of reactive networks?

Reactive networks are used in a variety of applications, including power factor correction, filters, and tuning circuits. They are also essential in the design of electronic devices such as radios, televisions, and computers.

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