Greetings,

I hope you are well, and I'd like to ask you about the power dissipated due to capacitor and inductor,

For example, a capacitor, say 2μF connected to the independent votage source 20V, the the source is removed and replaced by a resistor in such a way that the voltage does not change instantenously, the resistor has a resistance of 5Ω. and the qusetion is find the time when the power dissiapted is 75% of the initial power?

And what will be the time if we use inductor 2mH and do the process?

 

On a resistor, the power dissipaded is \(\frac{V^2}{R}\). Also, the voltage on the resistor will be described by the equation \(V_0 e^{-\frac{t}{RC}}\).
Therefore, the total power dissipated will be \(\frac{V_0 \cdot RC}{R}\) (someone correct me if I 'm wrong).
Then you need to solve the integral from 0 to t and equate that to 3/4 of \(\frac{V_0 \cdot RC}{R}\). Solving for t will give the answer.

 

^^

Thank you,

but when I saw solution in the the book, the author uses 1/3 rather than what you used. I don't why he solves like.

My question know, why he uses the remaining presntage of energy to calculate the time needed to dissiapate 75% ?

and In case of inductor, he uses the presnage itself not the remaining

 

As I posted, it is possible that I am wrong. I have never seen an exercise like yours so far. I put my logic down and solved it. Maybe if you write down what your book asks and suggests we will sort it out.

 

Perhaps the question is really asking at what time the stored energy in the capacitor has reduced to 75% of its original value - rather than when the power has fallen to 75%.

The stored energy at time t will be given by

\(W=\frac{1}{2}CV_c^2(t)\)

For the energy to have fallen by 75% would require the voltage to have fallen to √0.75 of the original value or 86.6% at time t.