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alexandr402 [8]
3 years ago
9

(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca

pacitor been connected to a 6.00-V battery, how much energy would have been stored?________ μJ
Physics
1 answer:
gulaghasi [49]3 years ago
6 0

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

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