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.

- anonymous

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- anonymous

@Michele_Laino

- 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

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)"

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## More answers

- anonymous

also, from what I know from kirchoffs voltage law V-Vr-Vc=0, so 2Vr=V when Vr=Vc

- anonymous

I think that relates to what my friend said.

- Michele_Laino

I think that the circuit can be modeled by this equation:
\[\frac{{dV}}{{dt}} + \frac{1}{{RC}}V = 0\]

- anonymous

Where does the V/RC come from?

- 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

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

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

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

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

where does 1/(wC)^2 come from?

- 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

\[\Large {Z_C} = \frac{1}{{\omega C}}\]

- 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

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

would V=I*(R+j/wC) apply to the resistor or the capacitor in this case?

- anonymous

either way the equation should be the same since Vr=Vc, right?

- 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

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

and since w=2pi*f, f=-j/(2piRC)?

- anonymous

Would this be the resulting answer?

- anonymous

or am i thinking to simplistically..

- 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

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

(RI being Vr as you mentioned of course)

- 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

w=sqrt(1/(3(CR)^2))?

- 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

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

Give me a moment to process. Sorry its pretty late here. 3:30 AM

- Michele_Laino

ok!

- Michele_Laino

in our case we have:
\[\Large {z_1} = \bar I,\quad {z_2} = R - \frac{j}{{\omega C}}\]

- Michele_Laino

so, the final result is:
\[\Large f = \frac{1}{{\sqrt 3 }}\frac{1}{{2\pi RC}}\]
so, you are correct!

- anonymous

From here, what should we do if we want to calculate the power of the resistor and the power of the capacitor?

- anonymous

sorry i meant the average power in the capacitor

- 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

this would be the power of the resistor with frequency f right?

- Michele_Laino

correct!

- anonymous

so then in our case we can simplify this equation to power (W) = 1/2(VI)?

- 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

Great! from here then can the same equation be applied to solve for the average power in the capacitor?

- 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

Oh so then after we plug things in and simplify, we would get that W = VI/(2sqrt(3)R^2) right?

- Michele_Laino

I got this:
\[W = VI\sin \varphi = \frac{{\sqrt 3 }}{2}VI\]

- Michele_Laino

since:
\[\sin \varphi = \sqrt {1 - {{\left( {\cos \varphi } \right)}^2}} \]

- 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

thanks!! :)

- 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 \(\sqrt {R^2+(\dfrac {1}{C\omega})^2}=2R\)

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