Engineering Electrical Circuits Reactive Networks Question

AI Thread Summary
At low frequencies, capacitors act as open circuits, preventing current from reaching the load, resulting in zero voltage. At high frequencies, inductors become open circuits, allowing current to flow through resistors and enabling voltage calculations using a voltage divider. The discussion highlights confusion regarding the impact of angular frequency on voltage and power calculations. The concept of bridged T filters is introduced, suggesting a potential transfer function approach for analyzing the circuit. Overall, the participant seeks clarification on their understanding and acknowledges gaps in their knowledge regarding certain circuit behaviors.
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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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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Because of the short circuits were all the inductors use to be, thus the resistors and the load are isolated
 
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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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
 
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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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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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
 

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