A capacitor circuit problem

Calculate the energy stored in a 1000 uF capacitor at t=50 us if the voltage across it is 1.5cos \[{10}^{5}\]t volts.

Satyaki. said:

Calculate the energy stored in a 1000 uF capacitor at t=50 us if the voltage across it is 1.5cos \[{10}^{5}\]t volts.

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Seems like it is your home work, right?

No friend. Its just a question I came across in Hayt & Kemmerly book. But my answer didnt match with the given one..

Satyaki. said:

Calculate the energy stored in a 1000 uF capacitor at t=50 us if the voltage across it is 1.5cos \[{10}^{5}\]t volts.

Click to expand...

This is a very strange way to specify a sinusoidal signal. Normally the argument (t) is either degrees or radians. But here it is specified as

(100000 * t)

So what are we to assume about this cos() function? If we assume the standard math theory interpretation that the argument is in radians, and if we assume that t is time in seconds, then we can re-write the argument as

(2 * PI) * (100000 / (2 * PI) ) * t

In terms of frequency in Hz, the cos function is cos( 2 * PI * freq * t)

So it looks like the signal specified in the problem has a frequency of 100000 / (2 * PI) = 15915.5 Hz. Furthermore, the problem seems to use the cos function merely as a roundabout means of specifying a fixed instantaneous charge. It starts out looking like a problem in time-varying signals, but ends up being a DC problem. I doubt that a problem like this teaches anything worthwhile.

Yeah ! U r right. The freq. Is that value.
The problem simply asks to find the energy=1/2cv^2 substituting the given capacitance and voltage values in the places of c and v respectively.

My question is whether we just put t=50 us sec in the above formula to find out energy or do we need to integrate it from 0 to 50 us to calculate energy.

But answer seems not to be matching either way. So, I was just wondering if there's some other roundabout to this sum.

The given ans is - 90.52 uJ

Satyaki. said:

Yeah ! U r right. The freq. Is that value.
The problem simply asks to find the energy=1/2cv^2 substituting the given capacitance and voltage values in the places of c and v respectively.

My question is whether we just put t=50 us sec in the above formula to find out energy or do we need to integrate it from 0 to 50 us to calculate energy.

But answer seems not to be matching either way. So, I was just wondering if there's some other roundabout to this sum.

The given ans is - 90.52 uJ

Click to expand...

At a given voltage on the capacitor, the capacitor has a given amount of charge. That charge implies a given amount of energy. No integration is needed. The time history of how the voltage got to that value does not matter. The only thing that matters is what that voltage is at t=50 sec.

Satyaki. said:

Yeah ! U r right. The freq. Is that value.
The problem simply asks to find the energy=1/2cv^2 substituting the given capacitance and voltage values in the places of c and v respectively.

My question is whether we just put t=50 us sec in the above formula to find out energy or do we need to integrate it from 0 to 50 us to calculate energy.

But answer seems not to be matching either way. So, I was just wondering if there's some other roundabout to this sum.

The given ans is - 90.52 uJ

Click to expand...

Just apply the equation you have E=1/2CV^2. I think you didn't get the correct answer because you either have the paranthese in the wrong place or maybe you are not in radians.
E=(1/2)*(1000uF)*((1.5*cos(100000 *50us))^2)

Remember only the voltage is squared. So V=1.5cos(100000*t) where t=50us. Once you find V, square it.

that should give you 90.52uJ. Remember you need to be in radians!

I think the answer is wrong. Its not coming !

I did exactly what you said !