A student places a capacitor across the terminals of this power supply. how does this produces a constant voltage?

Really don't get. I understand that charge builds up on the terminals of the capacitor, but how does that produce a constant voltage when the voltage from power supply varies?

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• 6 years ago

The equation for the voltage of a charged capacitor Vc(t) = Vco(e^-t/RC) which means that the voltage the cap C is charged to = Vco. It will discharge when connected to resistance R. Since R*C divides t in the exponent on e in the equation above, the larger R*C is, the longer it takes the cap C to discharge which means the longer it's voltage takes to fall from Vco. Think of it like a bathtub. The larger R*C is, the smaller the drain and the longer it takes to empty the bathtub. If R*C is small, Vc(t) will quickly fall and the filtering will be poor. Thus the larger C is, the better the filter.

• 6 years ago

Unless you are told otherwise, I would assume it is an ideal DC voltage source, so it has a constant voltage.

Does the problem explicitly say it is an AC source?

• 6 years ago

Suppose the power supply voltage is initially at V0 and you connect the capacitor. The capacitor will charge to V0 and hold a charge Q0 = CV0. Now since it is charged to V0, no current is flowing - essentially the capacitor acts as an open circuit across the voltage source.

If the voltage changes, say to V1, then the capacitor will charge up to V1 (if V1 > V0) or discharge to V1 (if V1 < V0) and current will flow in the process. Once the potential across capacitor reaches V1 then the current flow stops.

Suppose now you have a DC power supply but it has some AC "noise" - that is there are small fluctuations in voltage around the DC voltage, V0. Suppose now the capacitor and supply are both in parallel and you connect a load across both of them. When the voltage from the supply drops from V0, the capacitor discharges current to the load and assuming a constant impedance for the load, the voltage across the load does not change. If the voltage rises above V0, the capacitor charges, drawing some current away form the load so that the load gets the same current that is would from the volatge V0.