circuit solving - problem

[noname!]

delete

[aneevuser]

I guess you want to model the diode as either fully off or fully on with a 0.6 V diode drop across it?

If so, then I would draw a "diode on" and "diode off" equivalent circuit, and analyse those separately.

Note that you can just add the the DC and AC signals to compute the voltage that the diode feels at any time, and from that you'll be able to see when the diode is forward or reverse biased. There is no current flow until the diode is forward biased by 0.6 V, so no voltage drop across the 100R until then. When the diode is on, the current through it does not affect the voltage drop across it (it's always 0.6 V in our assumption) so you can apply KVL in the usual way to find i(t), and hence V_R(t) and U_D(t).

[noname!]

I think I have to assume that diode break down voltage is Ut, so it's 0V in this case. In order to compute the voltage i compute the rms so its 10sin(wt) => 10/sqrt(2)  = 7,07V. After that i add this voltage to DC, 7,07 + 5 = 12,07V. Right now i have total voltage AC + DC. If diode's break down is 0 V so Ud is always same as voltage drop on R, right? But i dont have idea how to compute current with KVL in order to compute max voltage drop and draw the diode's model.

[SG-1]

When adding AC + DC to get the RMS value.

Square each value, add together & take the square root of the sum.

Square root of (7.07²+5²)=8.66

I do not know if you are interested in the RMS value.

[rstofer]

LTspice - just to see what is happening.

[Zero999]

delete

No don't do that. It seriously pisses people off. Don't waste people's time.

[Simon]

delete

No don't do that. It seriously pisses people off. Don't waste people's time.

Yes and doing that will get one banned.