Answer :
Answer:52.98 V
Explanation:
Given
[tex]C=780 \mu F\approx 780\times 10^{-6}F[/tex]
[tex]L=0.290 H[/tex]
Current [tex]i=2.4 A[/tex]
[tex]\frac{\mathrm{d} i}{\mathrm{d} t}=89 A/s[/tex]
Now voltage across inductor[tex]=L\frac{\mathrm{d} i}{\mathrm{d} t}[/tex]
[tex]v=0.29\times 89=25.81 V[/tex]
same voltage is around the capacitor as they are in a loop
Total Energy[tex]=\frac{cV^2}{2}+\frac{Li^2}{2}[/tex]
[tex]E=\frac{780\times 10^{-6}\times 25.81^2}{2}+\frac{0.29\times 2.4^2}{2}[/tex]
[tex]E=0.2598+0.8352=1.095 J[/tex]
For maximum Voltage
[tex]\frac{cV_{max}^2}{2}=E[/tex] (as Energy is constant)
[tex]\frac{780\times 10^{-6}\times V_{max}^2}{2}=1.095[/tex]
[tex]V_{max}^2=2.807\times 10^3[/tex]
[tex]V_{max}=52.98 V[/tex]