Question

a 20-f capacitor is charged to 200 v. its stored energy is​

Answer 1

Explanation:

Here, V = 200 vol tq=0.1C

Energy relased on discharging = energy stored on charging

Answer 2

Answer:

The Stored energy inside the capacitor would be 0.4 Joule. Therefore, the correct answer would be 0.4 Joule.

Explanation:

Given: Capacitance of the capacitor, which is equal to = 20 μF = 20×10⁻⁶F

           Voltage up to which it is charged, which is equal to = 200 V

To Find: Energy stored in the capacitor.

Now, to find the energy stored in the capacitor, we know to know about the formula used to find the energy. There are basically three of them, that comprise of Q = charge stored, V = voltage applied and C = capacitance.

They are :

                                     $E$ = $\frac{CV^{2} }{2}$ = $\frac{QV}{2}$ = $\frac{Q^{2} }{2C}$

In the question, we have V and C. Hence we'll go with the first equation. It will become them :

                                      $E$ = $\frac{(20*10^{-6})(200*200) }{2}$

                                      $E = 400000 * 10^{-6}$

                                      E = 0.4 Joule

Therefore, the correct answer would be 0.4 Joule.

To know about virtual voltage : https://brainly.in/question/8239916

To know about breakdown voltage : https://brainly.in/question/17353578