Topic 11.3 – Capacitance – AHL Formative Assessment
NAME: _________________________________ TEAM:__
THIS IS A PRACTICE ASSESSMENT. Show formulas, substitutions, answers (in spaces provided) and units!
1. A 3.25-V battery is used to fully charge a 725 mF capacitor. How much charge was transferred from
the negative to the positive plate? 1. 0.002356 C
q = CV
= 72510-63.25
= 0.002356 C
Three 725 mF capacitors are connected in parallel to a 3.25 V battery.
2. What is the equivalent capacitance? 2. 0.002175 F
Cparallel = C1 + C2 + C3
= 3(72510-6)
= 0.002175 F
3. What is the charge on each capacitor? 3. 0.002356 C
Capacitors are in parallel, voltages are all the same and equal to 3.25 V.
q = CV
= 72510-63.25
= 0.002356 C
Three 725 mF capacitors are connected in series to a 3.25 V battery.
4. What is the equivalent capacitance? 4. 2.416710-4 F
1 / Cseries = 1 / C1 + 1 / C2 + 1 / C3
= 3(1 / 72510-6)
= 4137.931
Cseries = 1 /4137.931
= 2.416710-4 F
5. What is the voltage on each capacitor? 5. 1.083 v
In series voltages must add up to the battery voltage- so each capacitor has 3.25 / 3 = 1.083 v across its
plates.