# Question about AC and DC on a RC circuit

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?

• 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. – Chu Jul 2 '18 at 12:53

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.

• On behalf of educators everywhere, honest students everywhere, and EE stackexchange readers, please don't turn this into a homework answering service. – Elliot Alderson Jul 2 '18 at 10:49
• @ElliotAlderson You imagine incorrectly about the source of SpaceDog's question as well as this answer providing any homework answers for the OP's question. Take a close look at the phrasing as well as SpaceDog's other questions. SpaceDog merely needed to realize that this problem requires good familiarity with calculus. – jonk Jul 2 '18 at 10:56
• Yes, this is not a homework and even if it was, EE is here to explain what is asked, homework or not. If someone uses EE to do their homework the problem is with that someone, that will not learn. My problem here is not exactly calculus but realizing the correct situation so I can apply the correct approach. Your answer was accepted, sir. Thanks – Duck Jul 2 '18 at 16:00
• @ElliotAlderson SpaceDog has addressed himself to your accusation. Each of us has an equal right to choose our path and to earn such respect as others see fit. Just as a note towards another implication of yours, I used to teach as a professor at the largest university in my state. I think I've earned my own opinion on how to teach, or not to. You may not share it. Or respect it. But it's mine and I own it unabashedly. Best wishes. – jonk Jul 2 '18 at 17:47

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.

• why do you say that the situation is neither transient nor steady state? – Duck Jul 2 '18 at 9:25
• A transient situation has a stimulus that is a step function...closing or opening a switch in a circuit with dc sources. A steady state situation assumes that a sinusoidal stimulus has been present (essentially) forever. This problem is neither of those. – Elliot Alderson Jul 2 '18 at 10:48
• I don't understand because on my question I say that at t=0 I close the switch... – Duck Jul 2 '18 at 12:58
• But the switch does not connect a dc source so there is no step function stimulus. In a transient analysis the stimulus makes a single step change at t=0 and we analyze what happens after that...your source is changing continuously. – Elliot Alderson Jul 2 '18 at 13:30

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.