Question about AC and DC on a RC circuit

Question

Suppose a simple RC circuit, R and C in series with a voltage source and an open switch. The capacitor is discharged.

The power supply is AC, sinusoid with amplitude 10 V and a frequency of 50Hz.

So, the power supply is

10 Sin(100πt)

The resistor is 100Ω and the capacitor is 1F.

I turn the switch on at t=0.

What is the current at t=0.05 seconds?

As far as I see the current on the circuit is ruled by this formula:

so,

i(t) = (10 Sin(100πt)/100) e^(-0.05/100)

therefore, plugging t

i(t) = 27.05 mA

Now let use Ohm's law:

The circuit's impedance, Z is equal to

Z = R -j/wC

Z = 100 - j(2π)

Z = 100 - 2πj

so i = v/z

i = (10 Sin(100πt))/(100 -2πj)

i = 2.7/(100 -2πj)

How are these two values equal?

I was expecting to get the same values for both methods of calculation.

What am I missing?

Answer 1

It looks like you are mixing transient response analysis (the exponential equation) and steady-state ac analysis (the j\$\omega\$ part). In a situation like this, which is neither transient nor steady-state, I think you go back to basics: $$i_c = C \frac{dv_c}{dt}$$ and $$v_c = \cfrac{1}{C}\int i_c\, {dt} + v_c(0)$$

You also know that \$i_c = i_r\$ and \$V_c = V_s - R\,\cdot i_R\$. Looks like some math ahead.

Answer 2

I assume the schematic is this one:

^{simulate this circuit – Schematic created using CircuitLab}

The KCL is:

$$\frac{V_{\text{x}\left(t\right)}}{R_1}+C_1\:\frac{\text{d}\,V_{\text{x}\left(t\right)}}{\text{d} t}=\frac{V_{\text{1}\left(t\right)}}{R_1}$$

Both \$V_x\$ and \$V_1\$ are functions of time, with \$V_1\left(t\right)=V_\text{PK}\cdot\operatorname{sin}\left(\omega\: t\right)=10\cdot\operatorname{sin}\left(100\pi\: t\right)\$.

In standard form for first order linear differential equations, as used by first year calculus students to solve such equations, the above becomes:

$$\frac{\text{d}\,V_{\text{x}\left(t\right)}}{\text{d} t}+\frac{1}{R_1\:C_1}\:V_{\text{x}\left(t\right)}=\frac{V_{\text{1}\left(t\right)}}{R_1\:C_1}$$

Setting \$\tau=R_1\:C_1\$, the integrating factor is \$e^{\:t / \tau}\$ and so,

$$\begin{align*} \frac{\text{d}}{\text{d} t}\:\left(V_{\text{x}\left(t\right)}\:e^{\:t / \tau}\right)&=\frac{V_{\text{1}\left(t\right)}}{\tau}\:e^{\:t / \tau}\\\\ V_{\text{x}\left(t\right)}\:e^{\:t / \tau}&=\int \frac{V_{\text{1}\left(t\right)}}{\tau}\:e^{\:t / \tau}\:\text{d} t\\\\ V_{\text{x}\left(t\right)}&=e^{\:-t / \tau}\int \frac{V_{\text{1}\left(t\right)}}{\tau}\:e^{\:t / \tau}\:\text{d} t \end{align*}$$

Taking into account the initial condition that \$V_x\left(t=0\right)=0\:\text{V}\$ (in order to solve for the constant of integration, above):

$$\begin{align*}V_{\text{x}\left(t\right)}&=V_\text{PK}\cdot\frac{\omega\:\tau\left[e^{\:-t/\tau}-\operatorname{cos}\left(\omega\: t\right)\right]+\operatorname{sin}\left(\omega\: t\right)}{1+\omega^2\: \tau^2}\\\\&=V_\text{PK}\cdot\left[\frac{\omega\:\tau}{1+\omega^2\:\tau^2}\cdot e^{\:-t/\tau}+\frac{\operatorname{sin}\left(\omega\:t+\operatorname{tan}^{-1}\left(-\omega\:\tau\right)\right)}{\sqrt{1+\omega^2\:\tau^2}}\right]\end{align*}$$

From here, with \$V_\text{PK}=10\:\text{V}\$ and \$\omega=100\:\pi\:\frac{\text{rad}}{\text{s}}\$ and \$\tau=100\:\text{s}\$ and using \$I_{\left(t\right)}=\frac{V_{\text{1}\left(t\right)}-V_{\text{x}\left(t\right)}}{R_1}\$ at \$t=50\:\text{ms}\$. I get a value of \$-6.365\:\mu\text{A}\$.

---

Running a Spice deck:

v1 n001 0 sine(0 10 50)
r1 n001 vx 100
c1 vx 0 1
.tran 0 50m 0 10n uic
.meas TRAN CURRENT FIND I(R1) WHEN time=50m CROSS=1
.end

Spice reports:

current: i(r1)=-6.36461e-006 at 0.05

I think that means I probably didn't mess up on the equation solution.

Answer 3

The time constant is huge (100 sec) compared with the period of the applied voltage (0.02 sec), hence the transient can be ignored. At t=0.05 sec, the input has gone through 2.5 cycles, hence the voltage across R is 0V and the current is also zero.