anonymous
  • anonymous
A circuit consists of a sinusoidal voltage source, a resistor, and a capacitor in series. At what frequency will the voltage drop across the resistor be 1/2 of the voltage source? I have no idea how to start.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
Theres are 2 other parts to this question. Also, I'm sorry if I don't understand the concepts very well. I got sick for a week, and our lecturer covered quite a bit of material :(.
anonymous
  • anonymous
I know V=V0cos(wt), and from what I've heard from my friend "set the voltage drop across the resistor equal to the drop across the capacitor, so IR=I*(-i/wC)"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
also, from what I know from kirchoffs voltage law V-Vr-Vc=0, so 2Vr=V when Vr=Vc
anonymous
  • anonymous
I think that relates to what my friend said.
Michele_Laino
  • Michele_Laino
I think that the circuit can be modeled by this equation: \[\frac{{dV}}{{dt}} + \frac{1}{{RC}}V = 0\]
anonymous
  • anonymous
Where does the V/RC come from?
Michele_Laino
  • Michele_Laino
by the continuity equation: \[\Large \nabla \cdot {\mathbf{j}} + \frac{{\partial \rho }}{{\partial t}} = 0\] I can write: \[\Large I = - \frac{{dQ}}{{dt}}\] where \( \large I \) is the current of your circuit
anonymous
  • anonymous
If im not mistaken, that equation can be rearrange the previous equation dV/dt+V/RC=0 into C*dV/dt+V/R=0. This is then the sum of the currents in the resistor and the capacitor, but since it is in a closed loop, wouldn't the currents be the same throughout?
anonymous
  • anonymous
also, im not sure if it helps, but I hear people have been getting imaginary numbers for the end result for the frequency
Michele_Laino
  • Michele_Laino
Now I substitute that current formula above into this equation: \(\Large V=RI\): \[\Large V = - R\frac{{dQ}}{{dt}},\quad Q = CV\] yes you are right! with my equation above I refer to a transient state.
Michele_Laino
  • Michele_Laino
when we are out from the transient state, we can write these equations: \[\Large V = I\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} \]
anonymous
  • anonymous
where does 1/(wC)^2 come from?
Michele_Laino
  • Michele_Laino
It is the electric impedance of capacitor. \(\large RI\) is the voltage across the resistor, and : \[\Large \frac{I}{{\omega C}}\] is the voltage across the capacitor
Michele_Laino
  • Michele_Laino
\[\Large {Z_C} = \frac{1}{{\omega C}}\]
Michele_Laino
  • Michele_Laino
using complex numbers, the circuit, after the transient phase, can be modeled by this equation: \[\Large \bar V = \bar I\left( {R + \frac{j}{{\omega C}}} \right)\] where \(j^2=-1\)
anonymous
  • anonymous
I think im a little confused with where all these equations are coming from. Additionally, I dont have a full understanding of impedance. I believe impedance is just an obstruction to current, but I can't relate these equations.
anonymous
  • anonymous
would V=I*(R+j/wC) apply to the resistor or the capacitor in this case?
anonymous
  • anonymous
either way the equation should be the same since Vr=Vc, right?
Michele_Laino
  • Michele_Laino
yes! please I have made a typo, here is the right equation: \[\Large \bar V = \bar I\left( {R - \frac{j}{{\omega C}}} \right)\]
anonymous
  • anonymous
so when Vr=Vc, IR=I*(-j/wC) I think. From this, you can solve for w? so -j/wC=R and therefore w=-j/RC?
anonymous
  • anonymous
and since w=2pi*f, f=-j/(2piRC)?
anonymous
  • anonymous
Would this be the resulting answer?
anonymous
  • anonymous
or am i thinking to simplistically..
Michele_Laino
  • Michele_Laino
please, wait, the quantity: \[\bar I\left( {R - \frac{j}{{\omega C}}} \right)\] is a complex number, also current \(I\) is a complex number Since voltage across the resistor is: \(V=RI\) and voltage across the capacitor is \(V=I/(\omega\,C)\), then we can write: \[\Large \frac{{RI}}{{I\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} }} = \frac{1}{2}\] please solve for \(\Large \omega\)
anonymous
  • anonymous
Before I solve for w, the numerator, as you mentioned is RI. And since it is equal to 1/2, it can only mean that it is a ratio. Would this mean that the denominator of the left side is the voltage source? If not, how did you get the denominator of the left side? :o
anonymous
  • anonymous
(RI being Vr as you mentioned of course)
Michele_Laino
  • Michele_Laino
correct! the power source has to provide this voltage: \[\Large V = I\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} \]
anonymous
  • anonymous
w=sqrt(1/(3(CR)^2))?
Michele_Laino
  • Michele_Laino
I start from this equation: \[\Large \bar V = \bar I\left( {R - \frac{j}{{\omega C}}} \right)\] then I take the modulus of the complex numbers at the left and right side, so I can write: \[\Large {V^2} = {I^2}\left( {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} \right)\] finally I take the square root, so I get: \[\Large V = I\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} \]
Michele_Laino
  • Michele_Laino
I have used this identity: \[\Large \left| {{z_1}{z_2}} \right|^2 = \left| {{z_1}} \right|^2\left| {{z_2}} \right|^2\] where \(z_1,z_2\) are complex numbers
anonymous
  • anonymous
Give me a moment to process. Sorry its pretty late here. 3:30 AM
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
in our case we have: \[\Large {z_1} = \bar I,\quad {z_2} = R - \frac{j}{{\omega C}}\]
Michele_Laino
  • Michele_Laino
so, the final result is: \[\Large f = \frac{1}{{\sqrt 3 }}\frac{1}{{2\pi RC}}\] so, you are correct!
anonymous
  • anonymous
From here, what should we do if we want to calculate the power of the resistor and the power of the capacitor?
anonymous
  • anonymous
sorry i meant the average power in the capacitor
Michele_Laino
  • Michele_Laino
in order to compute the power \(W\), we can apply this formula: \[\Large W = VI\cos \varphi \] where: \[\Large \cos \varphi = \frac{R}{{\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} }}\] and \(V,I\) are the voltage and current respectively of the circuit
anonymous
  • anonymous
this would be the power of the resistor with frequency f right?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
so then in our case we can simplify this equation to power (W) = 1/2(VI)?
Michele_Laino
  • Michele_Laino
yes! at the frequency: \[f = \frac{1}{{\sqrt 3 }}\frac{1}{{2\pi RC}}\] the power provided by generator is: \[W = VI\cos \varphi = \frac{{VI}}{2}\]
anonymous
  • anonymous
Great! from here then can the same equation be applied to solve for the average power in the capacitor?
Michele_Laino
  • Michele_Laino
no, since we have to compute this: \[\Large W = VI\sin \varphi = VI\frac{{1/\omega C}}{{\sqrt {{R^2} + \frac{1}{{{{\left( {\omega C} \right)}^2}}}} }}\]
anonymous
  • anonymous
Oh so then after we plug things in and simplify, we would get that W = VI/(2sqrt(3)R^2) right?
Michele_Laino
  • Michele_Laino
I got this: \[W = VI\sin \varphi = \frac{{\sqrt 3 }}{2}VI\]
Michele_Laino
  • Michele_Laino
since: \[\sin \varphi = \sqrt {1 - {{\left( {\cos \varphi } \right)}^2}} \]
anonymous
  • anonymous
I found my mistake and got the same thing! Thank you so much for your help! Its rare to find someone so willing to help and amazing at physics! :D
Michele_Laino
  • Michele_Laino
thanks!! :)
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Since this is a series circuit, all components have the same intensity in common. Their voltages are proportional to their impedances. Hence, voltage across the resistor is proportional to its impedance \(R\); voltage acroos the generator is proportional to the impedance of the whole circuit, \(\sqrt {R^2+(\dfrac {1}{C\omega})^2}\) Solve \(\sqrt {R^2+(\dfrac {1}{C\omega})^2}=2R\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.