[tex] \boxed{ \bold{E = \frac{1}{2} CV {}^{2} }}[/tex]
where,
C = capacitance
V = voltage
C = [tex]300 μF = 300 \times {10}^{ - 6} F[/tex]
V = [tex]450 \: V[/tex]
E = ?
[tex]E = \frac{1}{2} (300 \times {10}^{ - 6} F)(450 \: V) {}^{2} [/tex]
[tex]E = 30.375 \: J[/tex]
[tex]E = 30.4 \: J[/tex]
[tex] \underline{ \underline{E = 30.4 \: J}}[/tex]
[tex]\: [/tex]
//@smokerz//
An energy storage capacitor is rated at 300 μF and 450 V. How much energy can it store?
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Verified answer
Formula:
[tex] \boxed{ \bold{E = \frac{1}{2} CV {}^{2} }}[/tex]
where,
C = capacitance
V = voltage
Given:
C = [tex]300 μF = 300 \times {10}^{ - 6} F[/tex]
V = [tex]450 \: V[/tex]
E = ?
Solution:
[tex]E = \frac{1}{2} (300 \times {10}^{ - 6} F)(450 \: V) {}^{2} [/tex]
[tex]E = 30.375 \: J[/tex]
[tex]E = 30.4 \: J[/tex]
Answer:
[tex] \underline{ \underline{E = 30.4 \: J}}[/tex]
[tex]\: [/tex]
//@smokerz//
An energy storage capacitor is rated at 300 μF and 450 V. How much energy can it store?
E = 30.4 J