Into capacitor, half wave rectifier diode conducts for?

[jaus tail]

Homework Statement

The time for which the diode conducts is

Homework Equations

Using integration and differentiation Vc = 1 / C integral (current)(dt)

The Attempt at a Solution

Vs = -Vc
Vm sin(wt) = -Vc = - (1/C) integral (Idt)
now differentiate both sides to get expression for I that is current.
Vm (coswt) * w = -(1/C) I
So I = Vm(costwt) * C * w (-1)

So now i draw current wave, and then how to proceed? How do you get an answer for this? Am i to find out when current goes negative?

There was a similar question for RL circuit that was

Vs = L di/dt
so integrate to get expression for current.
Vs sin(wt) = L di/dt
Vs sin (wt) dt = L di
integrating
Vs cos (wt) (-1) / w + k = L * I
I = Vs cos (wt) (-1) /(wL) + k' where k' = k/L equation 1

to find value of K' take initial condition. At t = 0, I = 0
so 0 = Vs (1) (-1) / (wL) + k' from equation 1

K' = Vs/(wL) equation 2

so expression for current becomes from equation 1 and 2

I = Vs cos(wt) (-1) / (wL) + Vs / (wL) = Vs/ (wL) * (1 - cos (wt))

Drawing this we get I is always positive, so it conducts for 360 degrees when input goes 1 entire cycle.
360 degrees was right for RL circuit but for RC answer is not given as the book is old and the page is torn.

i'm not sure what to do in case of given question of diode + C with charged as shown.
Expression is
I = Vm(costwt) * C * w (-1)

So what to now?

 

[NascentOxygen]

For the circuit shown, if the capacitor is initially uncharged then the diode will conduct for the entire first ¼-cycle of sinusoid V.sin ωt. After that, the diode will never conduct again.

BUT your thread title says "RC" so shouldn't you have included a resistance in parallel to the capacitor?

 

[jaus tail]

Thanks, Quarter means 45 degrees right?
I was unable to find how to change title but found now. Learnt 2 new things today.

 

[NascentOxygen]

Quarter means 45 degrees right?

A full cycle is 360°, a half-cycle is 180°, so a quarter-cycle is

 

[NascentOxygen]

I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.

 

[jaus tail]

I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.

Quarter cycle is 90 degrees. Darn i am so stupid.

We do need maths to find equation of current by integrating... Vc = 1/C (integration) (Idt)
I = Vm(costwt) * C * w (-1)

This is current waveform. So it'll flow from 0 to 90 degrees.

Thanks.

 

[CWatters]

I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.

+1

Usually there is a load in parallel with the capacitor that causes it to discharge. This in turn causes the diode to conduct for part of each cycle (after the initial 1/4 cycle). In fact problems like this usually ignore the initial 1/4 cycle and only want you to calculate the time per cycle after that.