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Capacitance and electric potential

A 2.61 micro F capacitor is charged to 1380 V and a 7.79 micro F is charged to 490 V. These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other.
What will be the charge on the 2.61 micro F capacitor and what will be the charge on the 7.79 micro F capacitor?
What will be the final potential difference across their assembly?

Solution Preview

When capacitors are connected in parallel, effective capacitance = C1+C2 = 2.61+7.79 = 10.4 micro ...

Solution Summary

Solution starts with computing the effective capacitance when capacitors are connected in parallel.