You have two batteries connected in series. The positive terminal of one battery is connected to a light bulb

and then to the positive terminal of the capacitor. The other battery is connected through a switch, to the negative terminal of the capacitor. One DMM (measures voltage) is connected across the capacitor (in series with the capacitor) while the other DMM is connected across the light bulb. The question is: How does the voltage change with time for the capacitor and for the light bulb?You have two batteries connected in series. The positive terminal of one battery is connected to a light bulbThe voltage across the capacitor will increase with the following function after the switch is closed:



Uc = U batt * 1-e^(-t/tc)



The voltage across the lamp is:



U batt - Uc



Uc = voltage across the capacitor.

U batt = battery voltage

t = time in seconds

tc = time constant = capacitance in Farads multiplied by resistance of lamp in Ohms (in seconds)



In the first moment after switching on, the lamp has the full voltage and the capacitor has zero voltage.

After some time the lamp has zero voltage and the capacitor is charged to the full battery voltage.