Help with circuits-reactive networks question

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In summary: Make sure to convert your resistance values to kΩ first. And double check your calculations for the time constant and voltage across R9. In summary, the conversation includes a discussion about a homework question, where the formula for energy stored is rearranged to find voltage and a voltage divider equation is used to find the voltage across a circuit. There is also a mention of finding the time constant and determining the voltage across a resistor using the voltage divider formula. Ultimately, the correct calculations are needed to find the answers.
  • #1
Davidlong
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Hi guys,

I have this homework question that i have attached. I've done a couple questions but I'm not sure if they are right.



For the first question i arrived at 15.49 by rearranging the formula; energy stored=1/2.C.V^2. when rearranged you get the square root of 240 for voltage.

For the second question b I don't understand how you find V. Is this the same as Vs? OR do you have to find the equivalent circuit by having all the inductors behave as a short circuit and capacitors as an open circuit since the circuit has reached a steady state.

For the third question I think i need the answer from b to continue.

For the fourth question i used the formula of power=V^2/Resistance. So 15.49^2/92 = 2.608mW

Not sure if this is right though.

Any help would be appreciated.

Thanks

 
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  • #2
Here's the attachment. The first question should be 21.91 using the formula from the last post.
 

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  • #3
Can you past the actual questions? Can't tell the players without a program.
 
  • #4
Sorry i uploaded the wrong file.

For the second question i got 109.54 by breaking the circuit down, replacing the capacitors with an open circuit and inductors with a short circuit then using the voltage divider equation.
v0 x (R1+R2/R2) = V.
 

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  • #5
Davidlong said:
Sorry i uploaded the wrong file.

For the second question i got 109.54 by breaking the circuit down, replacing the capacitors with an open circuit and inductors with a short circuit then using the voltage divider equation.
v0 x (R1+R2/R2) = V.

Yes, that looks reasonable. The result looks okay to me.
 
  • #6
For the 3rd question i used the formula V=Vo e^(-t/RC) and i got 0.00957 seconds for t. Does this seem correct as its very small.

For the fourth question i used the formula P=V^2/R, 21.91^2/92=5.21739, then i multiplied it by 1000 to give the answer in mW. Did i use the right numbers for this question?
 
  • #7
Davidlong said:
For the 3rd question i used the formula V=Vo e^(-t/RC) and i got 0.00957 seconds for t. Does this seem correct as its very small.
Show your work. How did you determine the time constant ##\tau = RC##?
For the fourth question i used the formula P=V^2/R, 21.91^2/92=5.21739, then i multiplied it by 1000 to give the answer in mW. Did i use the right numbers for this question?
Nope. How did you determine the voltage across R9?
 
  • #8
For R i found the equivalent resistant. R8 and R9 in parallel so the resistance when they are combined is 46. Then this new resistor is in series with R7 so they are added together to give 138. C is 0.0001 farads as it says in question 1.

so time constant is 138 x 0.0001=0.0138
0.0138 x ln0.5 = -t = -0.00957.

For question 4 i managed to find the voltage of R9 using the voltage divider formula.

21.91 x 46/92+46 = 7.3 V
46 is the equivalent resistance of R8 and R9.

so for the power i do 7.3^2/92 = 579.71mW

Is that correct?
 
  • #9
Davidlong said:
For R i found the equivalent resistant. R8 and R9 in parallel so the resistance when they are combined is 46. Then this new resistor is in series with R7 so they are added together to give 138. C is 0.0001 farads as it says in question 1.

so time constant is 138 x 0.0001=0.0138

0.0138 x ln0.5 = -t = -0.00957.
Ah. The resistors are specified in kΩ (kilo ohms), but you've treated them as just ohms. The equivalent resistance should be 1000X what you've used :wink:
[/quote]

For question 4 i managed to find the voltage of R9 using the voltage divider formula.

21.91 x 46/92+46 = 7.3 V
46 is the equivalent resistance of R8 and R9.

so for the power i do 7.3^2/92 = 579.71mW

Is that correct?[/QUOTE]

The power calculation looks okay.
 

1. What is a reactive network?

A reactive network is a type of electrical circuit that contains reactive components, such as capacitors and inductors, in addition to resistive components. These reactive components can store and release energy, resulting in a time-varying response to an applied voltage or current.

2. How do I calculate the impedance of a reactive network?

The impedance of a reactive network can be calculated using Ohm's Law, which states that impedance (Z) is equal to the voltage (V) divided by the current (I). However, in a reactive network, the impedance is a complex value, meaning it has both a real and imaginary component. To calculate the total impedance, you must use the complex impedance formula, which takes into account the resistive, capacitive, and inductive components of the circuit.

3. What is the difference between a series and parallel reactive network?

In a series reactive network, the components are connected in a single path, so the current flowing through each component is the same. In contrast, in a parallel reactive network, the components are connected across the same voltage source, resulting in the same voltage drop across each component. The main difference between the two is how the components are connected, which affects the overall impedance and current flow in the circuit.

4. How does frequency affect a reactive network?

The frequency of the applied voltage or current can greatly impact the behavior of a reactive network. At certain frequencies, the reactive components can either enhance or reduce the overall impedance of the circuit. This phenomenon is known as resonance and is a crucial concept in understanding the behavior of reactive networks.

5. How do I analyze a reactive network in a circuit simulation program?

To analyze a reactive network in a circuit simulation program, you will need to input the circuit's components, including resistors, capacitors, and inductors, as well as their values. You can then run the simulation and observe the behavior of the circuit, such as the voltage and current at various points, to understand how the components interact and affect the circuit's overall performance.

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