# Converting phasor back into time domain

1. Nov 13, 2016

### CoolDude420

1. The problem statement, all variables and given/known data
Find Amplitude of Steady-state voltage across capacitor

2. Relevant equations

3. The attempt at a solution

I'm at the very last step of changing it back into the time domain to obtain my ampltiude. I was planning on getting it into the form a + bj, then using trigonometry and other formulae to get it into polar form from which I can get it into the form of Bsin(wt + theta), with B giving me my amplitude.

Im stuck with 8+ 8/20 j^-1. Which isnt in the form a+bj. Any ideas how I can try and get it into that form?

2. Nov 13, 2016

### Staff: Mentor

There is no specific frequency information contained in that complex number. The frequency of all voltages and currents will be that of the driving signal.

Your complex number yields an angle and an amplitude.

How to view $\dfrac 1j$? Multiply both numerator and denominator by $j$ and see where that leads.

BTW, $\frac 8 {1+j20}$ is not equal to what you say. Instead, multiply numerator and denominator by the complex conjugate of the denominator so you can end up with a real denominator.

3. Nov 13, 2016

### CoolDude420

The frequency comes from the initial voltage sinusoid of 10rads/s. Phasors don't have frequencies.

If I multiply above and below by j. I get:

4. Nov 13, 2016

### Staff: Mentor

Check your voltage division work. The numerator of the fraction should contain the impedance of the component that you want the voltage across and the denominator the sum of the two impedances.

Regarding your conversion to time domain, take your rectangular form and convert it to polar form (magnitude and angle). You then have the magnitude and phase angle for the time domain.

5. Nov 13, 2016

### CoolDude420

Ah.. My Voltage division is wrong. I accidently used the formula for resistors in parallel. oops

6. Nov 13, 2016

### Staff: Mentor

Your expression for V right back at the start is wrong. Pretend you have resistors and write the expression, then substitute the capacitor impedance for one of the resistances.

7. Nov 13, 2016

### Staff: Mentor

By the way, the following is definitely not true:

You can't split a denominator like that, even for real values. Consider an example:

$\frac{8}{6 + 2} \ne \frac{8}{6} + \frac{8}{2}$

8. Nov 13, 2016

### CoolDude420

Okay. why am i being so stupid today. Apologies for all these stupid mistakes. Let me redo the question.

9. Nov 13, 2016

### CoolDude420

Okay. So I got

Now. How do I transform that into the form a+bj.?

10. Nov 13, 2016

### Staff: Mentor

That's standard complex math: multiply the numerator and denominator by the complex conjugate of the denominator.