Search Results

The capacitor is discharged. The power supply is AC, sinusoid with amplitude 10 V and a frequency of 50Hz. So, the power supply is 10 Sin(100πt) The resistor is 100Ω and the capacitor is 1F. …

I am a relative newbie to electronics. Looking at the following schematics I see something strange: C1 and C2 at the input. I always knew that capacitors block DC signals and let AC pass. By having …

The reactance of the capacitor will decrease with frequency. …

Suppose this circuit of an audio amplifier: How is C calculated? I suppose C is there to block DC, so it is a high pass filter. If this is the case the formula for C would be something like C = …

We know that voltage across a capacitor is given by But there is this other formula How this second formula was deducted mathematically from the first one? or from another one? …

I am watching this video where the guy deducts the voltage across the capacitor during the transient phase. There is an integral, in the middle of the equations. … This is not just this guy, I am not aware of any of the capacitor/inductor formulas where the tutor considers them. Why is that? …

I am watching this video where the guy is deriving the equations for the step function of a series RC circuit. At some point he has the following equation: \$ \frac{dt}{RC} = \frac{dv}{v_s - v} …

In this video, the instructor claims current will be 90 degrees behind voltage inside the capacitor but not inside the resistor or the whole circuit. If that's true, I don't understand. …

So, the circuit is stabilized and the voltage across the capacitor is 40V. At t=0, the switch is closed. What is the voltage across the capacitor 3ms after the switch is closed. … Now I want to know the voltage across the capacitor 3ms after it starts discharging. …

To find the equation for the voltage across the capacitor, I apply KVL to the circuit and get this: \$ V_1(t) = Ri + L\frac{di}{dt} + V_C(t) \$ this later turns in to this: \$ \frac{V_1}{LC} = \frac … The only thing I can see is that the current on a capacitor is equal to \$ i_c = C \frac{dv}{dt} \$ If R, L and C are in series, the current is the same for the three. …

omega_d t) + K_2 \thinspace Sin(\omega_d t) ] \$ To find the coefficients of those equations I apply the two initial conditions: I solve the equations for t=0 and for the initial voltage across the capacitor … In the voltage equations I used the initial voltage across the capacitor and the derivative of voltage (current). Now I have the current equations. …

Solid curves are voltage across capacitor. Dashed curve = current. V1 is a pulse, 10 VDC amplitude, 1ms. What am I missing? …

The documentation about this is vague and conflicting but as far as I have researched you calculate these capacitors like this: C1 Several authors say to calculate this capacitor equal to the input impedance …