Final answer:
A 165 µF capacitor stores approximately 1.169 ·² joules of energy when subjected to a voltage of 119 V, calculated using the energy storage formula E = ½ CV².
Step-by-step explanation:
Calculating Energy Stored in a Capacitor
To find out how much energy is stored in a 165 µF capacitor when a voltage of 119 V is applied, you can use the formula for the energy stored in a capacitor:
E = ½ CV²
Where:
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- E is the energy stored (in joules)
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- C is the capacitance (in farads)
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- V is the voltage (in volts)
Plugging in the values, we get:
E = ½ × 165µF × (119V)²
Convert µF to F (1µF = 10¹¹F):
E = ½ × 165 × 10¹¹ F × (119V)²
Now calculate the energy:
E = ½ × 165 × 10¹¹ × 14161 V²
E = ½ × 165 × 10¹¹ × 14161
E = 0.5 × 165 × 10¹¹ × 14161
E = 1169422.5 ·¹ × 10¹¹ J
E = 1.169 ·² J (rounded to three significant figures)
Therefore, a capacitor of 165 µF stores approximately 1.169 ·² joules of energy when 119 volts are applied to it.