The topic of this problem is Energy Storage Elements and the problem is to find i(t) the current in the circuit shown below. So we have a circuit that is composed of a voltage source 5+3e to minus 250t And it has four capacitors associated with it as well. So we're looking for i(t). The best way to approach this problem is to find an equivalent capacitance for the capacitances that we have in the circuit and then represent those as single capacitance. Then we'd be able to find our i(t) through our well-known relationship for The capacitor, I(t) is equal to CDVDT. So let's do that. So the first thing we notice on this is that we have two, four micro [INAUDIBLE]. Capacitors which were in series with one another. If you combine those in series with one another, they combine to be a two microfarad capacitance. And then you have this equivalent to microferret capacitance in parallel with another two microfarad capacitance, which again gives us 2 + 2 4 microfarads for the combination of the three right most capacitors in our circuit. And so now we a source we have a 4 microfarads capacitor that we haven't worked with yet, and an equivalent 4 microfarads capacitor for these three capacitors that we've reduced to an equivalency. So now we can take the four microfarad capacitor that we have remaining and combining it with the equivalent four microfarad capacitor that we have from the three capacitors and we get two microfarads. So we can redraw this circuit so it has a source in it. And it has an equivalent capacitance to the series and parallel combinations of capacitors that we have above, that's 2 microfarad. The source remains the same, it's 5+3e-250t, that's the volts. And our current is still the current flowing out of that voltage source. So again we're going to use our IV characteristics for a capacitor to solve the problem where i (t) = C dv/dt. And our capacitance is 2 microfarads so we can put that in there. 2 x 10 to the -6 for our capacitance. And we're taking the derivative of the voltage and our voltage, in this case, is 5 + 3e to the -250t. So, if we do that, we take the derivative of this equation, we end with (2 x -6) for our coefficient and the derivative of this is 3x250 e- 250t, so luckily we end up with a current i of t, had to be combined with all our terms. When we open the current it is equal to 1.5 e to -250t. And that's in milliamps.