Calculating Total Energy Stored in Parallel Capacitors

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To calculate the total energy stored in parallel capacitors C1 (25 µF) and C2 (6.0 µF) charged at 150 V, the correct formula is energy stored = 1/2 C (ΔV)^2. The initial calculations for charge were Q1 = 3750 µC and Q2 = 900 µC, leading to an energy calculation of 348750 J. However, the error arose from not converting capacitance from microfarads to farads before performing calculations. The corrected approach involves dividing the capacitance values by one million, which will yield the accurate energy stored in the capacitors. Proper unit conversion is crucial for obtaining the correct result.
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Two capacitors, C1 = 25 µF and C2 = 6.0 µF, are connected in parallel and charged with a 150 V power supply.

(a) Calculate the total energy stored in two capacitors.

energy stored = 1/2C(deltaV)^2 or Q^2/2C

I been doing this and its marking me wrong.

first i found both charges which is

Q = VC

C1 = 3750 uC
C2 = 900 uC

then i did

3750^2/2(25) = 281250

900^2/2(6) = 67500

i add both up and get 348750 J

It is saying my answer is incorrect and i did this problem 3 different way getting the same number.

does this look like a system error or is my number really off.
 
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Simple error in the units.

You need to take into account that the units of capacitance given are MICRO farads. Do the sums again but this time dividing first by a million before you square the terms.
 
Thanks for the correction.
 

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