Answer :

The voltage applied to the 6.00 mF capacitor to store 432 mJ of energy is 12 volts.

To find the voltage applied to a capacitor, you can use the formula:

Energy (E) = (1/2) * C * V^2

Where:

E = Energy stored in the capacitor

C = Capacitance

V = Voltage applied to the capacitor

In this case, the energy stored in the capacitor (E) is given as 432 mJ (millijoules), and the capacitance (C) is given as 6.00 mF (millifarads).

Let's substitute the given values into the formula and solve for V:

432 mJ = (1/2) * 6.00 mF * V^2

First, let's convert the energy and capacitance to joules and farads:

432 mJ = 0.432 J

6.00 mF = 0.006 F

Now, we can rewrite the equation:

0.432 J = (1/2) * 0.006 F * V^2

Divide both sides of the equation by (1/2) * 0.006 F:

0.432 J / [(1/2) * 0.006 F] = V^2

Simplify the left side:

0.432 J / (0.003 F) = V^2

V^2 = 144 V^2

Take the square root of both sides to solve for V:

V = √(144 V^2)

V = 12 V

Learn more about capacitors at https://brainly.com/question/21851402

#SPJ11

Thanks for taking the time to read How much voltage was applied to a 6 00 mF capacitor if it stores 432mJ of energy. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada