Is this right?
How much energy is stored in the 50mF capacitor when Va-Vb= 22V?
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For the parrallel of top25uF and bottom25uF, I got Ceq=25+25=50uF
For the series left50uF and right50uF, I got Ceq=1/50uF+ 1/50uF= 25uF as my final answer
To calculate the energy stored in a capacitor, we can use the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor,
C is the capacitance value in Farads (F), and
V is the voltage across the capacitor.
In your given circuit diagram, you have a 50mF (microfarad) capacitor, and the voltage difference across the capacitor is 22V.
To solve this problem, we need to convert the capacitance value from microfarads to farads. Since 1 microfarad (uF) is equal to 1e-6 Farads (F), the 50mF capacitance is equal to 50e-6 F.
Using the formula for energy stored in a capacitor, we can calculate the energy as follows:
E = (1/2) * C * V^2
E = (1/2) * (50e-6 F) * (22V)^2
Now, let's calculate the energy:
E = (1/2) * (50e-6 F) * (22V)^2
E ≈ (0.5) * (50e-6 F) * (484 V^2)
E ≈ 0.0121 joules (J)
Therefore, the energy stored in the 50mF capacitor when Va-Vb = 22V is approximately 0.0121 joules (J).