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1. What charge is stored in a 180.0-μFcapacitor when 120.0 V is applied to it?

Sagot :

Minihalfatimaidn

Answer:

So capacitance is charge divided by voltage and we can multiply both sides by V to solve for Q. So Q—the charge stored in the capacitor—is the capacitance multiplied by the voltage. So it's 180 times 10 to the minus 6 farads times 120 volts which is 0.0216 coulombs.

Explanation:

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