Question

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

Answer 1

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)

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

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

$V_{max}=52.98 V$