I have the following problem:
A 5 μF capacitor is connected across a voltage source = 170sin(377t) V.
What is capacitive reactance?
Here is my approach (\(f\) is not given):
I know, to find capacitive reactance i use formula \(X_C=\frac{1}{2\cdot\pi\cdot f\cdot C}\)
So that would be \(\frac{1}{2\cdot\Pi\cdot 0Hz\cdot 5uF}\) ? I don't even need to touch \(V_s\)?

 

In generic terms a voltage source could also be written as

Vs = Vp sin(ωt)

 

Reactance is for AC voltage, therefore will have a frequency. 0hz will not work. Both inductors and capacitors will behave similar to a resistor at various frequencies, The capacitor offers less reactivity at higher frequency, and the inductor prefers lower frequencies. Things like radios, remotes and speakers use this theory to only respond to specific frequencies. When the capacitative reactance is equal to the inductive reactance, the circuit is said to be at resonance. The 377t seems like it might be time?? perhaps there is a frequency there...

http://www.allaboutcircuits.com/vol_2/chpt_3/2.html

 

Did it escape your notice that

2∏*60 ≈ 377

So your capacitive reactance is

1 / (377)(5e-6) = ? Ohms

BTW (1/0) would approach infinity so that cannot be the answer

 

That's good insight papa, I didn't look into the 377t, We were always given the frequency. I've never seen it written with a "t". I was thinking the t was for time, and 1/t = f.

 

You need to use the angular frequency, ω which from the given equation is equal to 377.

Dont forget ω = 2 ∏f

 

Papabravo, thanks for clearing me on this that \(f=60Hz\)
\(X_C=\frac{1}{377\cdot 5uF}=1.8kohms\)

On the same problem (A 5 μF capacitor is connected across a voltage source = 170Vsin(377t) what is the equation of current and there are multiple answers:

a. i = 0.32 sin(377t+90°)A
b. i = 0.32 sin(377t-90°)A
c. i = 0.5 sin(377t+90°)A
d. i = 0.5 sin(377t-90°)A
e. i = 0.5 sin(377t+180°)

But I don't get any of those answers. Since \(I=\frac{V}{R}\)
So \(I=\frac{170Vsin(377t)}{1.8kohms}=94.4V\)
What am I doing wrong?

 

I have the following problem:
A 5 μF capacitor is connected across a voltage source = 170sin(377t) V.
What is capacitive reactance?
Here is my approach (\(f\) is not given):
I know, to find capacitive reactance i use formula \(X_C=\frac{1}{2\cdot\pi\cdot f\cdot C}\)
So that would be \(\frac{1}{2\cdot\Pi\cdot 0Hz\cdot 5uF}\) ? I don't even need to touch \(V_s\)?

I'm unsure how you ended up with 0Hz for frequency.

170 is the amplitude, while 377t is frequency. The function will give you the amplitude and phase at time t

\(2 \cdot \pi \cdot \omega = f\)
\(Z_c = \frac{1}{\omega \cdot C\)

The 2π is added to your equation to get degrees per second instead of radians per second for frequency.

 

I'm unsure how you ended up with 0Hz for frequency.

170 is the amplitude, while 377t is frequency. The function will give you the amplitude and phase at time t

\(2 \cdot \pi \cdot \omega = f\)
\(Z_c = \frac{1}{\omega \cdot C\)

The 2π is added to your equation to get degrees per second instead of radians per second for frequency.

Thanks I already figured out that \(f=60Hz\), for some reason i didnt figure it out at first.

 

Papabravo, thanks for clearing me on this that \(f=60Hz\)
\(X_C=\frac{1}{377\cdot 5uF}=1.8kohms\)

On the same problem (A 5 μF capacitor is connected across a voltage source = 170Vsin(377t) what is the equation of current and there are multiple answers:

a. i = 0.32 sin(377t+90°)A
b. i = 0.32 sin(377t-90°)A
c. i = 0.5 sin(377t+90°)A
d. i = 0.5 sin(377t-90°)A
e. i = 0.5 sin(377t+180°)

But I don't get any of those answers. Since \(I=\frac{V}{R}\)
So \(I=\frac{170Vsin(377t)}{1.8kohms}=94.4V\)
What am I doing wrong?

Obviously calculation is not your strong suit. Remember a uF is one one-millionth of a Farad.

(377 * 5e-6)^-1 = 1 / (377 * 5 * 10^-6) = 1 / (.001885) ≈ 531 Ohms

The actual impedance of the capacitor is a complex number which is 531 Ohms at an angle of -90°. You could also write 0 - j531

so you would have

170*sin(377t + 0°) / (531, -90° )
≈ 0.32sin(377t + 90°)
That is 
{170 / 531, 0° - (-90°)} = {0.32, +90°}

which corresponds to answer (a.)

Remember ELI the ICE man which tells you that current I leads voltage V in a capacitor.