# Regarding reservoir capacitor in diode bridge [duplicate]

In my recent physics lab I constructed and examined the diode bridge (see https://en.wikipedia.org/wiki/Diode_bridge#Output_smoothing).

In particular I was interested in the smoothing properties of the capacitor. However when adding the capacitor ($300 \mu F$) and measuring the voltage (continously, using voltage probes over the resistor) I noticed that the resultant voltage curve was only smoothed by certain resistors.

This might be a basic question concerning the behaviour of current, but why will certain low-ohm resistors (I believe I used a 100 $\Omega$ resistor) result in a (unchanged) pulsating voltage while other high level resistor (for instance 10000 $\Omega$ resistor) creates the intended smoothing?

## 1 Answer

If you have a resistor of small resistance, than for a given voltage this will result with a big current flowing through the resistor. Capacitor smoothes output of the diode bridge by storing electric energy. When the output voltage of the diode bridge is falling, the capacitor acts a a temporary source of current. However if the output current is big, the capacitor will discharge quickly and there will be big fluctuations of the output voltage and current. In general, what determines the "smoothnes" of the output signal is the product of resistance and capacitance (RC time constant), which should be much bigger than one fourth of the alternating current frequency.