RC circuit

Discussion in 'Homework Help' started by reallylongnickname, Jul 5, 2012.

  1. reallylongnickname

    Thread Starter New Member

    Jul 5, 2012
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    [​IMG] Question asks, what is the voltage drop across R1?

    To solve, I see there is parallel and series. I suppose the impedance formula will work, \sqrt{R^2}+{Xc^2} but what about the parallel part?
     
    Last edited: Jul 5, 2012
  2. WBahn

    Moderator

    Mar 31, 2012
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    Q1) If you knew the current in R1, what would be the voltage drop across it?

    Q2) If you knew the current in the voltage source was Io, what would the current be in R1?

    Q3) What is the effective impedance (remember, need either both magnitude and phase angle or real and imaginary components) of the parallel combination of R2 and XC2?

    Q4) What is the effective impedance of the result from Q3 in series with R1 and XC1?

    Q5) What is the current, Io, in the voltage source (using the result from Q4 as the effective load)?
     
  3. reallylongnickname

    Thread Starter New Member

    Jul 5, 2012
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    z=/sqrt{180^2}+{240^2} = 300 ohms

    I =V/R
    I =10/300
    I =33mA

    So far so good?
     
  4. Engr

    Member

    Mar 17, 2010
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    5
    solving impedance in parallel is just like solving resistors in parallel but you need to take note of the phase angle of the impedance in parallel. if you can solve the total impedance then you can compute for the current passing through R1 then you can solve the voltage drop across R1.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Let's take a step back.

    What kind of math are you currently using in your analysis. Specifically, are you familiar with and using phasors? Are you familiar with and using complex numbers?
     
  6. reallylongnickname

    Thread Starter New Member

    Jul 5, 2012
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    phasors ya, complex numbers no. I have a separate formula for phasor angle in parallel.
    tan^-1\frac{R}{Xc}
     
    Last edited: Jul 5, 2012
  7. WBahn

    Moderator

    Mar 31, 2012
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    My condolences. Working with complex numbers makes everything SO much easier. But you work with what you've got.

    So, let's go back to my original questions and modify them slightly to relfect the use of phasors only. Please answer each question in turn; if nothing else, I need to get a handle on the terminology and conventions you are using.

    Q1) If you knew the current in R1, what would be the voltage drop across it?

    Q2) If you knew the current in the voltage source was Io, what would the current be in R1?

    Q3) What is the effective impedance (both magnitude and phase angle) of the parallel combination of R2 and XC2?

    Q4) What two components in series (a resistor in series with either an inductor or a capacitor) has the same impedance (both magnitude and phase) as the effective impedance from Q3?

    Q5) With the series circuit from Q4 replacing the parallel combination of R2 and XC2, you should now have the voltage source and four components in series - R1, XC1, and the two components from Q4. What is the effective two-component series circuit that is equivalent to all four so that you have a single resistor in series with a single reactive element (either a capacitor or an inductor)?

    Q6) What is the magnitude and phase angle of the two-component series circuit from Q5?

    Q7) What is the current, Io, in the voltage source (using the result from Q4 as the effective load)?
     
  8. reallylongnickname

    Thread Starter New Member

    Jul 5, 2012
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    Q1) v = I*R = 33mA*180 ohms = 5.94v

    Q2) Io ? I don't know what this represents.

    Q3) z=\frac{180*240}{\sqrt180^2+240^2}= 144 ohms

    \left(\frac12{180}{240}\right)" alt="tan^-1\left(\frac12{180}{240}\right)" />
     
    Last edited: Jul 6, 2012
  9. reallylongnickname

    Thread Starter New Member

    Jul 5, 2012
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    Disregard the above post answer.

    Q1) v = I*R = 33mA*180 ohms = 5.94v

    Q2) Io ? I don't know what this represents.

    Q3) z= \frac{333*250}{\sqrt333^2+250^2} =200 ohms

     tan^-1* \frac{333}{250} = 53 degrees

    Q4) ?
     
    Last edited: Jul 6, 2012
  10. WBahn

    Moderator

    Mar 31, 2012
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    Just v= I*R.

    The I is an unknown. Remember, we are starting from square one. Don't assume that your 33mA is correct (it isn't).

    The questions stated that it represents the current in the supply. Whatever the current is in the supply, that is what Io is equal to. It is simply a symbolic variable. How much algebra have you had so far.

    The thing I was trying to get you to recognize here is that if we can determine Io, the current in the supply, then we know what the current in the 180Ω resistor is because they are in series and hence have the same current.

    It should be -53.1 degrees (as a rule, carry three sig figs unless there is a good reason to do otherwise. For intermediate results, it is reasonable to carry at least one, but seldom more than two additional sig figs to minimize propagation of round off errors).

    Go back and look at your material and tell us what it says about the phasor angles associated with a capacitor and an inductor. Usually, capacitors are -90deg and inductors are +90deg. But, then again, the capacitive reactance, X_C, is generally negative and the inductive reactance, X_L, is positive. So that is at odds with the reactance of the two capacitors being given as positive values. So I would like to know how you have been taught to deal with this.
     
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