We constructed the circuit below

and when measuring the output we got a ripple factor of abut 25%.


We then added a 10 ohm resistor next to the capacitor as shown,

and the results were a ripple factor of 29%.


I've been stuck and can't figure out why the ripple factor went up just by adding a small resistor in series with the capacitor.

 

Why would you think the resistor would have no effect?
The load is drawing current from the capacitor during the time it's not being charged from the diode.
This current has to flow through the resistor in series with the capacitor which causes a voltage drop across the resistor.
This voltage drop subtracts from the capacitor output voltage so the capacitor voltage is lower than without the resistor.
If you view the resistor voltage you will see this.

 

Also, it slows the rate at which the capacitor can be recharged.

 

...
If you view the resistor voltage you will see this.

Plot the capacitor current with and without the resistor.