Unknown circuit in a black box

[Lambda96]

Homework Statement

What does the circuit look like and what are the values of resistor $R$ and capacitance $C$?

Relevant Equations

none

Hi,

I am not sure if I have calculated the task correctly

I have now assumed that the capacitor does not need to be charged and is therefore fully charged. In a DC circuit, a capacitor acts like an infinitely large resistor or like an open switch, so I assumed that it is a parallel circuit and that it looks like this.

The resistor then has the following value $R=100 \Omega$.

Using the impedance and the value for $R$, I can then calculate the value for $C$.

The value for the impedance in parallel circuit is

$$|Z|=\frac{1}{\sqrt{R^-2+\omega^2 C^2}}$$

$C$ can then be calculated as follows

$$C=\frac{R^2-Z^2}{R \omega z}$$

If I now substitute all the values into the above formula, I get the following $C=9.7 \cdot 10^{-4}F$.

 

[SammyS]

ƒHomework Statement: What does the circuit look like and what are the values of resistor $R$ and capacitance $C$?
Relevant Equations: none

Hi,

I am not sure if I have calculated the task correctly

View attachment 327711
I have now assumed that the capacitor does not need to be charged and is therefore fully charged. In a DC circuit, a capacitor acts like an infinitely large resistor or like an open switch, so I assumed that it is a parallel circuit and that it looks like this.

View attachment 327712

The resistor then has the following value $R=100 \Omega$.

Using the impedance and the value for $R$, I can then calculate the value for $C$.

The value for the impedance in parallel circuit is

$$|Z|=\frac{1}{\sqrt{R^-2+\omega^2 C^2}}$$

$C$ can then be calculated as follows

$$C=\frac{R^2-Z^2}{R \omega z}$$

If I now substitute all the values into the above formula, I get the following $C=9.7 \cdot 10^{-4}F$.

It looks like you used frequency, ƒ, rather than angular frequency. Remember, $\omega=2\pi f$ .

LaTeX tip:
To write $R^{-2}$ rather than $R^-2$, place the $-2$ in braces { }, e.g. $\{ -2 \}$ .

 

[Lambda96]

Thanks SammyS for your help and for looking over my calculation, thanks also for the tip on how to write $R^{-2}$ in latex

You are right, unfortunately I used the frequency of $50Hz$ in the calculation instead of $\omega=2 \pi 50Hz$. Then the result for $C$ is as follows, $C=0.015 F$

 

[SammyS]

Thanks SammyS for your help and for looking over my calculation, thanks also for the tip on how to write $R^{-2}$ in latex

You are right, unfortunately I used the frequency of $50Hz$ in the calculation instead of $\omega=2 \pi 50Hz$. Then the result for $C$ is as follows, $C=0.015 F$

The decimal point is in the wrong place.

Simply divide your original answer by $2\pi$ .