Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V
C1= 3×10^-6 F
V1= 480v
C2= 4×10^-6 F
V2= 500v
(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V
Simplifying the above, we get:
( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.
Further simplified as:
3440 × 10^-6 = 7 × 10^-6 × V
Making V the subject
V = 491.43volts
Therefore the potential difference across each capacitor is 491.43v
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Answer:
the power in both cases is the same.
Explanation:
hope helps you
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1) the strong nuclear force, 2) the electromagnetic force, 3) the weak nuclear force, and 4) gravity
