When an A.C voltage source in series with the DC voltage source are applied to a capacitor in series with a resistor they say that capacitor will block Dc and will let AC pass to the resistor. I am not understating it. By using super position theorem the statement can be proved but i am not getting the concept. Since the voltage/current across the capacitor is now pulsating Dc in this case then how the capacitor is blocking DC while letting AC passing through the resistor? How and why a capacitor makes a pulsating DC input to an AC output across the resistor ?
Electrical – Capacitor with pulsating dc voltage source/current
accapacitordc
Related Solutions
Solving ckt#3 the hard way using differential equations:
To start with, this equations always holds, for any capacitor $$i = CdV/dt$$
In the circuit you've provided, we have two unknown voltages (V1 across C1 and V2 across C2). These can be solved by applying Kirchoff's Current Laws on the two nodes.
For node V1: $$ (V_s-V_1)/R_1 = C_1 dV_1/dt + (V_1-V_2)/R_2 $$
And for node V2: $$ (V_1-V_2)/R_2 = C_2 dV_2/dt $$
Now we've got two differential equations in two unknowns. Solving the two simultaneously give us the expressions for V1 and V2. Once V1 and V2 are calculated, calculating the currents through the branches is trivial.
Solving differential equations is, of course, not trivial. What we generally do is to use Laplace Transform or Fourier Transform to convert them into algebraic equations in the frequency domain, solve the unknowns, and then do Inverse Laplace/Fourier transform to get the unknowns back into time domain.
Method 2: Use voltage divider rule:
If we recall that the impedance across a capacitor C is $$Z=1/jwC$$ and denoting the impedances of the two capacitors C1 and C2 as Z1 and Z2, we can calculate V2 using the formula for voltage division across two impedances (http://en.wikipedia.org/wiki/Voltage_divider): $$V_2 = V_1 R_2/(R_2 + Z_2)$$ V1 can also be calculated using the same rule, the only issue is that the impedance on the right side of node 1 is a bit complex: it's the parallel combination of Z1 and (R2 + Z2). V1 now becomes $$V_1 = V_s (Z_1*(R_2+Z_2)/(Z_1+R_2+Z_2))/(R_1 + (Z_1*(R_2+Z_2)/(Z_1+R_2+Z_2)))$$
What to do next is to expand Z1 and Z2 using the capacitive-impedance formula, to get V1 and V2 in terms of w. If you need the complete time response of the variables, you can do Inverse Fourier Transforms and get V1 and V2 as functions of time. If however, you just the need the final (steady-state) value, you can set $$w=0$$ and evaluate V1 and V2.
A rather simpler way:
This method can give only the final steady-state values, but it's a bit handy for quick calculations. The catch is that once a circuit has settled into a steady state, the current through every capacitor will be zero. Take the first circuit (the simple RC) for example. The fact that the current through C is zero dictates the current through R (and hence the voltage drop across it) also to be zero. Hence, the voltage across C will be equal to Vs.
For the second circuit, all the current must pass through the path R1->R2->R3 if the capacitor draws no current. This means the voltage across C (equal to the voltage across R2) is $$V_s R_2 / (R_1 + R_2 + R_3)$$
In the last circuit, current through C2 being equal to zero implies the current through R2 being zero (and hence any voltage drop across it). This means any current that flows must take the path R1->C1. However, the current through C1 is also zero, which means R1 also carries no current. So both the voltages V1 and V2 will be equal to Vs in steady state.
The basic relationship in a capacitor is that the voltage is proportional to the charge on the "+" plate. However, we need to know how current and voltage are related. To derive that relationship you need to realize that the current flowing into the capacitor is the rate of charge flow into the capacitor. Here's the situation. We'll start with a capacitor with a time-varying voltage, v(t), defined across the capacitor, and a time-varying current, i(t), flowing into the capacitor. The current, i(t), flows into the "+" terminal taking the "+" terminal using the voltage polarity definition. Using this definition we have:
ic(t) = C dvc(t)/dt
This relationship is the fundamental relationship between current and voltage in a capacitor. It is not a simple proportional relationship like we found for a resistor. The derivative of voltage that appears in the expression for current means that we have to deal with calculus and differential equations here - whether we want to or not.
Best Answer
Maybe this image helps you understand:
From the time domain point of view you might think of this "pulsating DC" concept and think that it should pass over the capactior.
From the frequency domain point of view, there are two different components, the DC (0Hz) and AC. If you put a capacitor in series, you are using a High Pass Filter, meaning that the DC component will be filtered and won't pass.