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What is the voltage across this capacitor, inductor and resistor?

  • #1
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Homework Statement:

Circuit containing a 270ohm resistor in series with a 15uf capacitor 150mh inductor which are in paralell

Relevant Equations:

what is the voltage across the capacitor?
what is the voltage across the inductor?
what is the voltage across the resistor?
I can solve for the questions in completely series or parallel circuits however having the capacitor and inductor in parallel while the resistor stays in series is stumping me completely.
 

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Answers and Replies

  • #2
Baluncore
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You have a voltage divider with the resistor in series with the parallel LC.
How do you work out the combined resistance of two resistors parallel?
How do you work out the combined impedance of the L and C in parallel?
Remember that the impedance of a capacitor is negative and an inductor is positive.
 
  • #3
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It is helpful in these circuits to think of each these elements as a complex impedance:
Resistors: Z = R
Inductors: Z = jωL
Capacitors: Z = 1/(jωC)
( j = √-1, ω = 2πf )

Then solve this like you would for a resistor network, except the answer will be a complex number. For a problem like this, f = the frequency of the source. The magnitude of the complex result is the "amount" (Voltage, Current, etc.), the phase is the phase shift relative to whatever had zero phase in the problem statement, which is usually the source.
 
  • #4
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Thank you Dave,

At risk of sounding well understudied is there a way to solve this with simple algebra?
I was under the assumption there would be a variation on the existing equations used to find Z and X in simple RLC circuits which would eventually solve like a literal resistor network.
However after spending hours on this problem I am beginning to think I am in over my head, in all honesty I am not sure what to do with an imaginary number when it comes to circuits.
 
  • #5
168
1
You have a voltage divider with the resistor in series with the parallel LC.
How do you work out the combined resistance of two resistors parallel?
How do you work out the combined impedance of the L and C in parallel?
Remember that the impedance of a capacitor is negative and an inductor is positive.
Combined resistance would be using 1/((1/R1)+(1/R2)) which as far as I know works with impedance just like it does resistance. The reactance of L would be (2*pi*f*L) and the reactance of C would be (1/(2*pi*C))
Giving me:

R=270
Xc=24.1
Xl=414.7

I could find the impedance of C and L by taking Xc+R or Xl+R however I cannot figure out how to do this when they are sharing R.
 
  • #6
Baluncore
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Xc=24.1
Xl=414.7
but you have forgotten that Xc is always negative;
so Xc and XL cancel to some extent.
Xp = 1 / ( 1/Xc + 1/XL )
 
  • #7
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but you have forgotten that Xc is always negative;
so Xc and XL cancel to some extent.
Xp = 1 / ( 1/Xc + 1/XL )

Yes Yes! ok so now
1/(1/-24.1 + 1/414.7)
Xp=-25.6
Therefore current through R is Ir = (270-25.6) = 244.4
That doesn't sound right...
 
  • #8
Baluncore
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Now you must consider that the same current flows through the Rs and Xp, so the voltage across the reactance is at 90 deg to the voltage across the resistance, and they sum to the applied voltage. Plot reactive Vx on the y axis, and resistive Vr along the x-axis. Scale it so the diagonal has length proportional to the applied voltage.
 
  • #9
SammyS
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Combined resistance would be using 1/((1/R1)+(1/R2)) which as far as I know works with impedance just like it does resistance. The reactance of L would be (2*pi*f*L) and the reactance of C would be (1/(2*pi*C))
It's not quite so simple with impedances, unless you express them using complex numbers as in Dave's reply.

But for several reactances in parallel with no resistances involved, a rule holds which is similar to that for resistances in parallel.

In your case, ##\ \displaystyle X_\text{Eq.}=\frac{1}{\dfrac{1}{X_L}+\dfrac{1}{X_C}}
=\frac{X_L X_C}{X_L+X_C}\ \, \ ## as @Baluncore stated, Also, use ##X_C=\dfrac{-1}{\omega C} \ . ##

After that, be careful about how to add resistances and reactances in series.
 
  • #10
SammyS
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Yes Yes! ok so now
1/(1/-24.1 + 1/414.7)
Xp=-25.6
Therefore current through R is Ir = (270-25.6) = 244.4
That doesn't sound right...
... because it's not right.

The current through R is the same as the current through the parallel combination represented by ##X_p##

But as @Baluncore states, the voltages are 90° out of phase.

Using the hint from your screenshot, but using Vp rather than VL − VC you have ## V = \sqrt{V_p^2 + V_R^2} ##.

This crudely gives ## V = ZI = \sqrt{(X_pI)^2 + (RI)^2} = I \sqrt{(X_p)^2 + (R)^2}##.
 

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