# Archived A Resistor and a Capacitor in a Series AC Circuit

1. Nov 29, 2007

### sskakam

1. The problem statement, all variables and given/known data
A resistor with resistance R and a capacitor with capacitance C are connected in series to an AC voltage source. The time-dependent voltage across the capacitor is given by V_c(t) = V_c_0 sin(wt).

A) What is the amplitude I_0 of the total current I(t) in the circuit?

B and C will be revealed after A, so I may return after I get A.

2. Relevant equations
i = I cos(wt)
Vc = I/(wt)
V=IR

3. The attempt at a solution
V_c(t) = I(t)*R = (V_c_0)sin(wt)
Icos(wt)=((V_c_0)sin(wt))/R
I=((V_c_0)/R)tan(wt)

2. Feb 7, 2016

### Staff: Mentor

The magnitude of the current through a reactive component (inductor or capacitor) is given by Ohm's law using the component's reactance in place of resistance. Begin by determining the reactance of the capacitor:

$X_C = \frac{1}{ωC}$

Next, we are given the expression for the voltage across the capacitor as a sinewave of magnitude $V_{co}$. So the magnitude of the current will be:

$I = \frac{V_{co}}{X_C} = V_{co} ω C$

Another approach is to start with the relationship between current and voltage for a capacitor:

$I = C \frac{dv}{dt}$

We are given $V_c(t) = V_{co} sin(ω t)$ so that

$I(t) = C \frac{d}{dt}(V_{co} sin(ω t)) = V_{co}ωC cos(ω t)$

and the magnitude can be extracted from the expression for the current.

3. Feb 7, 2016

### CrazyNinja

@gneill .. there is a resistor in the circuit too. So we can invoke the concept of impedance. I don't know LaTeX that well and I believe you can do it better anyway.

4. Feb 7, 2016

### Staff: Mentor

Yes, an impedance approach would work very well. I avoided the use of impedance because the OP seemed to be using introductory level concepts, working with the time domain functions of voltage and current. I presumed he was just being introduced to reactance and working with that. Feel free to add your version.

We are specifically given the voltage across the capacitor, so that gives us the current through the capacitor regardless of what else is in the circuit. Since it's a series circuit, the rest of the components will have the same current. The situation is like this:

We aren't given the source voltage, but no doubt it'll have a phase angle associated with it if we're taking the capacitor signal as the reference!