# Need help for quiz related to circuit analysis

#### Manish55

Joined Apr 17, 2024
5
I came across below circuit analysis problem

So this is the situation during second positive half cycle. (Each cycle being 20 ms).
Conside the diode on right hand side. The voltage at anode of diode will be -10 V as it is connected to capacitor's negative plate.
What will be voltage at cathode of diode. During positive half cycle its cathode is connected to negative terminal of voltage source and positive plate of capacitor. Also whether capacitor will start discharging towards negative terminal of voltage source. Content creator was saying voltage across the diode will be -10 V. That means whether cathode is at group potential. Please somebody help me understand how did we arrive the voltage at the cathode and voltage across diode here

#### MrChips

Joined Oct 2, 2009
30,984
The lower diode is always reversed biased. Hence you can remove it from the circuit. The capacitor is charged to -10V and never gets discharged.

#### Manish55

Joined Apr 17, 2024
5
Why capacitor never gets discharged?

#### Manish55

Joined Apr 17, 2024
5
The lower diode is always reversed biased. Hence you can remove it from the circuit. The capacitor is charged to -10V and never gets discharged.
Why capacitor never gets discharged? It can get discharged right through negative terminal (during second positive half cycle)

#### MrChips

Joined Oct 2, 2009
30,984
On the negative half of the cycle, the upper diode is forward biased and the capacitor becomes charged.
On subsequent negative cycles, the voltage across the diode is zero and no further current flows. The capacitor is already charged to -10V.

On the positive half of the cycle, both diodes are reversed biased. No current flows through the diodes.

#### Manish55

Joined Apr 17, 2024
5
On the negative half of the cycle, the upper diode is forward biased and the capacitor becomes charged.
On subsequent negative cycles, the voltage across the diode is zero and no further current flows. The capacitor is already charged to -10V.

On the positive half of the cycle, both diodes are reversed biased. No current flows through the diodes.
Ok thanks
I understood most of ot when I watched the video again
Basically mu confusion is due to capacitor remains charged during positive cycle also and not getting discharged
Capacitors discharge depends on RC time constant. But there was no mention of it in the explanation in the video. Since diode become open circuited there was no way for capacitor to get discharged and that is why the capacitors remain charged through subsequent cycles after gettimg charged. Is this correct understanding
Also if we want to understand in terms of RC time constant, then probably
RC becomes infinite as diodes become open circuited due to reverse bias and offer infinite resistance as those are ideal diodes. Is this correct understanding

#### MrChips

Joined Oct 2, 2009
30,984
You are correct. Quite simply, once the capacitor is fully charged to -10V, neither diode will conduct. There is no path for current to flow.

#### Manish55

Joined Apr 17, 2024
5
You are correct. Quite simply, once the capacitor is fully charged to -10V, neither diode will conduct. There is no path for current to flow.
Thanks

#### MisterBill2

Joined Jan 23, 2018
19,042
For any voltage to follow an RC time constant there must be an "R" . Since there is no "R" to provide a path for current, there is no current.
AND,why add a ground connection to a circuit that does not relate to ground in any way at all???