The voltage and current at the terminals of the circuit...

The voltage and current at the terminals of the circuit element in (Figure 1) are zero for t < 0. For t ? 0 they are v = 70e^-1600t - 70e^-400t V, i = 4.0e^-1600t - 4.0e^-400t mA. Part B How much energy is delivered to the circuit element between 0 and 620 ?s? Express your answer to three significant figures and include the appropriate units. w = Part C Find the total energy delivered to the element. Express your answer to three significant figures and include the appropriate units. w_total =

Transcript

00:01 So let's say we're given the voltage in a circuit as a function of time is 70 times e to the minus 1 ,600 t, minus 70 times e to the minus 400t volts.

00:17 So all these units are in volts.

00:20 And they're also given the current as a function of time is 4 times e to the minus 1 ,600 t, and then minus 4 times e to the minus 400t, and this is in millie amps.

00:40 And part a asks, what is the power, or how much energy is delivered to the circuit in 620 microseconds? so delta t is 620 microseconds.

00:53 So the power is going to be the voltage times the current, and this is going to be a function of time, so these are not constant.

01:01 And so that means the energy delivered in a certain time interval is going to be the integral from zero to, we'll write this as 620 microseconds of the product of these two integrated with respect to time.

01:20 So if we do that, and the problem doesn't say what units t is in in the equation.

01:26 I'm going to assume seconds, but i could be wrong.

01:31 But anyway, so if we proceed with this, so i'll have.

01:34 Have we have a common factor of 70 we can factor out of the voltage terms so e to the minus 1600 t minus e to the minus 400 t and then this is times four millie amps or i guess we could say four milli watts because if the voltage is in volts and the currents in milli amps and that produces four milli watts and this is times e to the minus or it's basically the same exponential uh squared and this is integrated with respect to t.

02:12 So if we write it out, we'll have 280 milliwats is our sort of coefficient.

02:18 And then this is from 0 to 6 .2 times 10 to the negative 4 seconds.

02:24 And we'll have e, our first term will be e to the minus 3200 t, minus twice the cross term...

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