Question

A 4.0-μF capacitor that is initially uncharged is connected in series with a 4.0-kΩ resistor and an ideal 17.0-V battery. How much energy is stored in the capacitor 17 ms after the battery has been connected?

Answer 1

Answer:

2.473x10^-4J

Explanation:

C=(4x10^-6)farads, R=4000omhs, Vinital=17v, t=.017s

Use these equations τ=RC, Vfinial=Vinital(1-e^(-t/τ)), and En=$\frac{CV^2}{2}$

τ=(4x10^-6)(4000)=.016

Vfinial=17(1-e^(-.017/.016))=11.12v

En=$\frac{(4x10^-6)(11.12^2)}{2}$=2.473x10^-4J

Answer 2

The number of energy that should be stored in the capacitor should be 2.473x10^-4J

Calculation of the no of energy:

Since A 4.0-μF capacitor that is initially uncharged is connected in series with a 4.0-kΩ resistor and an ideal 17.0-V battery.

And,

τ=RC, Vfinial=Vinital(1-e^(-t/τ)), and En = CV^2/2

Now

τ=(4x10^-6)(4000)=.016

Vfinial=17(1-e^(-.017/.016))=11.12v

Now finally the energy should be

= (4x10^-6)(11.12^2) / 2

= 2.473x10^-4J

Hence, The number of energy that should be stored in the capacitor should be 2.473x10^-4J

Learn more about energy here: https://brainly.com/question/13024379