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

Step-by-step explanation:

Given

`C=780 \mu F\approx 780* 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* 89=25.81 V`

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

Total Energy
`=(cV^2)/(2)+(Li^2)/(2)`

`E=(780* 10^(-6)* 25.81^2)/(2)+(0.29* 2.4^2)/(2)`

`E=0.2598+0.8352=1.095 J`

For maximum Voltage

`(cV_(max)^2)/(2)=E` (as Energy is constant)

`(780* 10^(-6)* V_(max)^2)/(2)=1.095`

`V_(max)^2=2.807* 10^3`

`V_(max)=52.98 V`