• Since the square of the voltage appears in the energy formula, the energy stored is always positive. You can't have a negative amount of energy in the capacitor.  That means you can put energy into the capacitor, and you can take it out, but you can't take out more than you put in.

• Power in to the capacitor can be negative. Voltage can be positive while current is negative. Imagine a capacitor that is charged. You could charge a capacitor by putting a battery across the capacitor, for example. Then, if you placed a resistor across the capacitor, charge would leave the capacitor - current would flow out of the capacitor - and the energy in the capacitor would leave the capacitor only to become heat energy in the resistor. When energy leaves the capacitor, power is negative.

• When you use capacitors in a circuit and you analyze the circuit you need to be careful about sign conventions. Here are the conventions we used, and these conventions were assumed in any results we got in this lesson.

Frequency Dependent Behavior For A Capacitor

       We start with a capacitor with a sinusoidal voltage across it.

where:

• v_C(t) = Voltage across the capacitor

• i_C(t) = Current through the capacitor

• C = Capacitance (in farads)

        We will assume that the voltage across the capacitor is sinusoidal:

v_C(t) = V_max sin(wt)

        Knowing the voltage across the capacitor allows us to calculate the current:

i_C(t) = C dv_C(t)/dt = w

C V_max cos(wt) = I_max cos(wt)

      where  I_max = wC V_max

Comparing the expressions for the voltage and current we note the following.

• The voltage and the current are both sinusoidal signals (a sine function or a cosine function) at the same frequency.

• The current leads the voltage.  In other words, the peak of the current occurs earlier in time than the peak of the voltage signal.

• The current leads the voltage by exactly 90^o.  It will always be exactly 90^o in a capacitor.

• The magnitude of the current and the magnitude of the voltage are related:

        Now, with these observations in hand, it is possible to see that there may be an algebraic way to express all of these facts and relationships.  The method reduces to the following.

• If we have a circuit with sinusoidally varying voltages and currents (as we would have in a circuit with resistors, capacitors and inductors and sinusoidal voltage and current sources) we associate a complex variable with every voltage and current in the circuit.

• The complex variable for a voltage or current encodes the amplitude and phase for that    voltage or current.

• The voltage and current variables can be used (using complex algebra) to predict circuit behavior just as though the circuit were a resistive circuit.

        We need to do two things here.  First, we can illustrate what we mean with an example.  Secondly, we need to justify the claim above.  We will look at an example first, and we will do two examples.  The first example is jsut the capacitor - all by itself.  The second example will be one that you have considered earlier, a simple RC low-pass filter.

Example 1 - The Capacitor

        In a capacitor with sinusoidal voltage and currents, we have:

where:

• v_C(t) = Voltage across the inductor

• v_C(t) = V_max sin(wt)

• i_C(t) = Current through the inductor

• i_C(t) = wC V_max cos(wt) = I_max cos(wt)

• C = Capacitance (in farads)

        We represent the voltage with a complex variable, V.  Considering this as a complex variable, it has a magnitude of V_maxand and angle of 0^o.  We would write:

V = V_max/0^o

        Similarly, we can get a representation for the current.  However, first note:

i_C(t) = wC V_max cos(wt) = I_max cos(wt) = I_max sin(wt + 90^o)

(Here you must excuse the mixing of radians and degrees in the argument of the sine.  The only excuse is that everyone does it!)  Anyhow, we have:

I = I_max/90^o = j I_max = jwC V_max

Where j is the square root of -1.

        Then we would write:

V/I = V_max/jwC V_max = 1/jwC

and the quantity 1/jwC

is called the impedance of the capacitor.  In the next section we will apply that concept to a small circuit - one you should have seen before.

        Before moving to the next section, a little reflection is in order.  Here are some points to think about.

• A phasor summarizes information about a sinusoidal signal.  Magnitude and phase information are encoded into the phasor.  Frequency information is not encoded, and there is a tacit assumption that all signals are of the same frequency, which would be the case in a linear circuit with sinusoidal voltage and current sources.

• We looked at a case where we encoded a signal V_max sin(wt)

  into a phasor of V_max/0^o.  That was completely arbitrary, and many others would have encoded V_max cos(wt)

  into a phasor of V_max/0^o.

• Phasors are intended only to show relative phase information, and it doesn't matter which way you go.

Using Impedance

        In the last section we began to talk about the concept of impedance.  Let us do that a little more formally.  We begin by defining terms.

        A sinusoidally varying signal (v_C(t) = V_max sin(wt)

for example) will be represented by a phasor, V, that incorporates the magnitude and phase angle of the signal as a magnitude and angle in a complex number.  Examples include these taken from the last section.  (Note that these phasors have nothing to do with any TV program about outer space.)

is represented by a phasor V = V_max/0^o

is represented by a phasor I = I_max/90^o

is represented by a phasor V_a = V_A/f

        Next, we can use the relationships for voltage and current phasors to analyze a circuit.  Here is the circuit.

        Now, this circuit is really a frequency dependent voltage divider, and it is analyzed differently in another lesson.  However, here we will use phasors.  At the end of this analysis, you should compare how difficult it is using phasors to the method in the other lesson.

        We start by noting that the current in the circuit - and there is only one current - has a phasor representation:

We will use the current phase as a reference, and measure all other phases from the current's phase.  That's an arbitrary decision, but that's the way we will start.

        Next we note that we can compute the voltage across the capacitor.

This expression relates the current phasor to the phasor that represents the voltage across the capacitor.  The quantity 1/jwC is the impedance of the capacitor.  In the last section  we justified this relationship. We can also compute the phasor for the voltage across the resistor.

V_R = IR

This looks amazingly like Ohm's law, and it is, in fact, Ohm's law, but it is in phasor form.  For that matter, the relationship between voltage and current phasors in a capacitor - just above - may be considered a generalized form of Ohm's law!

        Now, we can also apply Kirchhoff's Voltage Law (KVL) to compute the phasor for the input voltage.

        You should note the similarities in what happens here and what happens when you have two resistors in series.

• If you have a resistor, R, and a capacitor, C, in series, the current phasor can be computed by dividing the input voltage phasor by the sum of R and 1/jwC.

• If you have two resistors in series (call them R₁and R₂), the current can be computed by dividing the input voltage by the sum of R₁and R₂.

Example

        Consider a series circuit of a resistor and capacitor.  The series impedance is:

That's the same as we showed just above.  The impedance can be used to predict relationships between voltage and current.  Assume that the voltage across the series connection is given by:

That corresponds to having a voltage phasor of:

We also know that the impedance establishes a relationship between the voltage and current phasors in the series circuit.  In particular, the voltage phasor is the product of the current phasor and the impedance.

For our particular impedance, we have:

So, we can solve for the current phasor:

Now, we know the voltage phasor and we know the impedance so we can compute the current phasor.  Let us look at some particular values.

Assume:

• R = 1.0 kW

• C = .1mf = 10^-7 f

• f = 1 kHz, so w= 2p10³

• V_max = 20 v

Then:

• Z_R = 1.0 kW

• Z_C = 1/(jwC) = 1/(j2p10³ 10^-7 ) = j 1.59 kW

And, the total impedance is:

• Z = Z_R + Z_C = (1.0 + j 1.59) kW

This impedance value can also be expressed in polar notation:

• Z =  1.878 /62^o

Now, compute the current phasor:

• I = V / (R + 1/jwC)

Substituting values, we find:

• I = V / Z = V_max/0^o / 1.878 /62^o =20/0^o / 1.878 /62^o

• I = V / Z = (20 / 1.878) /-62^o = 10.65 /-62^oamps

And, we need to examine exactly what this means for the current as a function of time.  But that isn't very difficult.  We can write out the expression for the current from what we have above.

• i_C(t) = 10.65 cos(wt - 62^o) amps