A capacitor C is charged to an initial potential of 60.0 V, with an initial char

Question

A capacitor C is charged to an initial potential of 60.0 V, with an initial charge of Q0. It is in a circuit with a switch and an inductor with inductance L = 3.87×10-2 H and resistance RL.

At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t.

Calculate the energy in the circuit after a time of 12 periods. Note that the curve passes through a grid intersection point.

Calculate the time required for 85% of the initial energy to be dissipated.

Explanation / Answer

The potential is decreasing contineouly as time is passing on. There are toal 11th periods are given, for 12th period analyzing the data , the potential will come down to 16volt.

The relationship between energy stored on capcitor plates and voltage is given as

E = 0.5 CV2 ..................1

Here C = ?

frequancy = 1/2pi (LC)1/2

C = 1 / 2pi f2 L

We need frequancy to find capacitance.once capacitance is found then we can find the value of energy using equation1. In order to find out the value of frequancy let us having a look upon the curve. Time taken for 12periods will be 12T =1.1ms

T = 0.091sec

Thus the relationship between frequancy f and T is

f = 1/ T = 1/ 0.091 = 10.91Hz

Putting the value of frequancy to get the value of C

C = 1 / 2(3.14)(3.87)(10-2)(10.9)2

   C = 0.0346F

Putting the v alue of C in equ1

E = 0.5(0.0346)(60)2

E =62.33J

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