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The first question is whether this is for a class or real life. In a class, this sort of "figure out whether the diode is off or on" problem usually assumes an ideal diode with zero forward voltage dr …

Phase is just a way of describing time differences in periodic signals and events. Remember that, from a mathematical standpoint, a true periodic signal is eternal: $$f(t + T) = f(t), \ \ \ -\infty …

In mathematical theory, "pure DC" is eternal -- it never begins or ends. A theoretical DC-powered oscillator must also be eternal. … But if the DC voltage was always there, and the oscillation was always happening, I'm not sure it makes sense to say that the oscillation comes from the DC source. …

Found it! In your schematic for the \$E_4\$ Thevenin voltage, you drew \$E_4\$ backwards. The positive end should point towards the 2k resistor. This led to a sign error that caused you to miscalculat …

Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop. If a loop has a branch that's not part of any other loop, that does guar …

Your diagram is more or less what happens. After connecting the battery, you end up with a net positive charge on the top (red) plate and a net negative charge on the bottom (blue) plate. There's an e …

condition: $$v_o = v_i - 0 \cdot e^{-t/RC} = v_i$$ So at DC, the capacitor acts like an open circuit. … This means that at DC, you can put a large voltage across a capacitor without current flowing through it. …

As DoxyLover pointed out, it's not just a matter of "getting hot". The efficiency of a linear regulator is Vout / Vin, which is really bad when there's a large difference between input and output. Co …

This means you can divide the solution into three steps: DC circuit analysis before the switching event (initial condition) DC circuit analysis a long time after the switching event (final condition) …

You're getting exactly what one would expect from that circuit. You haven't made a switch, you've made a source follower (aka a common drain amplifier). We had another question along these lines recen …

Hint #1: You need to work with the Thevenin equivalent looking into nodes A and B from \$R_P\$. Hint #2: What is the condition for maximum power transfer? (\$R_L =\ ?\$) Hint #3: You know the total …

Short version: Pulsed DC is actually AC. *The charge and discharge are actually exponential decays, so mathematically, the current never really stops. …

If the circuit changes, it's not a DC circuit anymore (meaning #1) A circuit that contains only DC sources (meaning #2) and passive components (resistors, capacitors, and inductors) will eventually (asymptotically … ) become a DC circuit (meaning #1). …

I'm not an electrician, but I can help with the circuit analysis. The formula for resistance is: $$R = \rho \frac L A$$ where \$\rho\$ is the resistivity of the steel, \$L\$ is the length of the wire, …

Something is wrong here. The range between the two supplies in 4.4V, but with a 212.6 μA current, the drop across the source and drain resistors is: $$212.6\mathrm{\mu A} \cdot (6000\mathrm{\Omega} + …