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Lera25 [3.4K]
3 years ago
10

A charged capacitor with C = 780 μF is connected in parallel to an inductor that has L = 0.290 H and negligible resistance. At a

n instant when the current in the inductor is i = 2.40 A , the current is increasing at a rate of di/dt=89.0A/s.what is the maximum voltage across the capacitor?
Physics
1 answer:
lora16 [44]3 years ago
3 0

Answer:52.98 V

Explanation:

Given

C=780 \mu F\approx 780\times 10^{-6}F

L=0.290 H

Current i=2.4 A

\frac{\mathrm{d} i}{\mathrm{d} t}=89 A/s

Now voltage across inductor=L\frac{\mathrm{d} i}{\mathrm{d} t}

v=0.29\times 89=25.81 V

same voltage is around the capacitor as they are in a loop

Total Energy=\frac{cV^2}{2}+\frac{Li^2}{2}

E=\frac{780\times 10^{-6}\times 25.81^2}{2}+\frac{0.29\times 2.4^2}{2}

E=0.2598+0.8352=1.095 J

For maximum Voltage

\frac{cV_{max}^2}{2}=E   (as Energy is constant)

\frac{780\times 10^{-6}\times V_{max}^2}{2}=1.095

V_{max}^2=2.807\times 10^3

V_{max}=52.98 V

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