Recent content by Potatoishere

Thanks a lot guys for the great explanation, I finally got it :)

During which the capacitor is discharging, will the current passing through the resistor be the sum of current from the capacitor and the AC source?

When capacitor is discharging, there is no current coming from the AC source at all? Once the AC source has reached a maximum voltage, the current will start to decrease but there will still be current coming out from the AC source into load, am I right?

Why would the diode stop conducting when the pd across the capacitor is greater than the AC source?

Let's assume two batteries and a resistor connected in parallel, and assuming one battery has higher emf than another(), what will be the voltage across the load? I think this scenario is very much similar to full-wave rectifier with smoothing. When the capacitor is discharging, it is because...

What about the voltage from the AC generator when capacitor is discharging?

From the graph when the capacitor is discharging, why is the voltage drop across the load is the equal to change in voltage across the capacitor? Since there is still current coming out from AC generator, hence, shouldn't the voltage drop across the load equal to the sum of voltage drop across...

So when the capacitor is discharging, will the there still be current coming from the AC generator?

I am sorry, but I still can't see why the capacitor can't discharge through the generator from the diode diagram? I think my main trouble is that I can't seem to comprehend that during the first half cycle, when the AC generator reaches its peak voltage, the capacitor will also be charged up at...

Even though the capacitor is in parallel with the resistor, but there are still current coming from the AC generator. And hence, wouldn't the sum of current through the resistor be the current from the capacitor and the generator? And V=IR, thus the voltage drop across the resistor must be the...

So in a full wave rectification, a capacitor is being connected in parallel with a resistor. During the period of capacitor discharging, will the capacitor discharge through the AC generator aswell? And why is the voltage across the resistor is equal to the change of voltage in the capacitor?

I got it. Thank you for the great explanation! :smile::thumbup:

so for example, in an artificial nuclear fission; let's take Uranium 235 as an example. As U235 being bombarded by a moving neutron, it splits out two smaller nuclei Br+Kr+3neutrons and energy is released in the process. So it is right to think that K.E of the neutron is being totally converted...

To be able split the larger nucleus into smaller nuclei, don't you need to supply external work onto the system? And my curiosity is since binding energy is the energy needed to break the bond of the deuteron apart, therefore external work must be applied onto the system, so my question is since...

Can someone please explain why is the difference in binding energy is equal to the energy released in a nuclear fission/fusion. From my understanding of binding energy, it is the energy needed to break bond between neutron and proton or energy released when forming into a nucleus. However, what...