A/C circuits Pre-Lab questions

[Amcote]

Homework Statement

For the RC circuit shown in Fig. 1, at some frequency the peak voltage across the capacitor and resistor are equal. Find the frequency at which this occurs. Show that the peak voltage across the capacitor or the resistor at this frequency is given by V_in/ √ 2. How would this compare if we had two resistors R instead of a resistor and a capacitor?

fig 1. is an RC circuit with 1 capacitor (0.1μF), 1 resistor (1kΩ) and a voltage source.

Homework Equations

1. Z_c = −i/ ωC
2. |Z_c| = |V_c|/ |I| = 1/ ωC ,
3. R = V/I
4. ω=2πf

The Attempt at a Solution

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To be honest it's been a very long time since I've done any physics and I'm having a little difficulty re-learning all the circuit stuff.

To get the frequency the only thing I can think of is setting V_c equal to V to get

IR=I/wC=I/2πfC

Then,

f=1/2πCR

so,

f=1/2π(0.1μF)(1kΩ)=1591.55Hz

As for the next part, I'm really not sure what it is asking. Perhaps if I knew exactly what V_in/ √ 2 is then I'd be able to make some progress.

But using the formulas, the peak voltage across the capacitor is

|V_c|/ |I| = 1/ ωC (I think)

And I don't know a formula for the peak voltage across the resistor, the only formula I have for the resistor is

R = V/IAnd actually another thing I am confused about is the difference between resistance and impedance because it seems they use these terms interchangeably. Thanks guys.

 

[Simon Bridge]

Is this a series or parallel circuit?
If V is the voltage across the resistor, then putting V_c=V is just what the description says.
Your reasoning looks long winded: if $V_c=V$ then $Z_c=$?

The next part is asking you to rewrite your equation for the frequency in terms of the input voltage. Clearly the frequency cannot be the (scaled) input voltage since they don't have the same units.

Impedence is resistance that depends on frequency... so, at a constant frequency, they are the same thing.

 

[Amcote]

This is a series circuit.

So what I have so far is that we are trying to find the frequency when |V_R|=|V_C|

|V_R|=IR
|V_C=I(1/wC)

IR=I(1/wC), w=2πf

Rearrange to get
f=1/(2πCR), or w=1/(CR)

Now to show that the peak voltage for the cap and the resistor at this frequency can be written as V_in/√2

I can take the ratio |V_C|/|V_in|= (1/wC)/(R²+(1/wC)²)^1/2

and so
|V_C|/|V_in|=1/ (1+(wCR)²)^1/2

plug in w=1/(CR) (from before)

and I get

|V_C|/|V_in|=1/ (1+(CR/CR)²)^1/2=|V_C|/|V_in|=1/ (1+(1)²)^1/2=1/√2

thus |V_C|=|V_in|/√2

 

[Simon Bridge]

Well done.