# Time averages and Phasors

1. Jul 20, 2009

### jeff1evesque

Current and Phasors

Statement:
A simple circuit is given, such that the AC power source "V" is in parallel with a resistor "G" (why would it be denoted by G instead of R?), and also parallel with a capacitor "C".

Relevant equations:
Converting the circuit to phasors, the current from the source is given by:
$$I = GV + C\frac{dV}{dt} = GV + j\omega CV$$ (#1)

I know $$I = \frac{V}{R}, C\frac{dV}{dt}$$ are equations of ohm's law for a resistor and capacitor respectively. I also know by Kirchhoff's law the sum of the

"branches" is equal to the main source.

My Question:
I am not sure what the symbol "G" denotes (in my notes I wrote that G = 1/R), nor sure how to get the second equality in equation (#1). I wrote in my notes that GV is a

constant so there is no derivative of it. But how do we take the phasor form (and derivative?) of $$GV + C\frac{dV}{dt}$$

thank you,

JL

Last edited: Jul 20, 2009
2. Jul 20, 2009

### CEL

Re: Current and Phasors

G is the notation for conductance, the inverse of resistance. Ohm's law can be written
V = RI
or
I = GV.
GV is not a constant, since V is variable. To derive GV, follow the rule of derivation of a function (V) multiplied by a constant (G).

3. Jul 20, 2009

### jeff1evesque

Re: Current and Phasors

I almost understand, but I still don't know how the phasor form from my question.

Last edited: Jul 20, 2009
4. Jul 20, 2009

### CEL

Re: Current and Phasors

The current GV is in phase with the voltage. The curent $$j \omega CV$$ is 90 degrees out of phase with the voltage.

5. Jul 20, 2009

### jeff1evesque

I'm guessing thats the same as writing $$\frac{Vcos(\omega t + \theta - 90\circ)}{\frac{1}{\omega C}} = \frac{Vsin(\omega t + \theta)}{\frac{1}{\omega C}}?$$

Thanks

Last edited: Jul 20, 2009
6. Jul 20, 2009

### jeff1evesque

But why is there no imaginary "j" term in the notation above to match $$j\omega CV?$$. I've derived the equation above by assuming the power source is given by the equation $$Vcos(\omega t)$$. Since the current though a capacitor is defined by $$C\frac{dV}{dt}$$, then we can write $$C\frac{d(Vcos(\omega t))}{dt} = \frac{Vsin(\omega t + \theta)}{\frac{1}{\omega C}}}.$$

Last edited: Jul 20, 2009
7. Jul 20, 2009

### CEL

Because $$j\omega$$ corresponds to $$\frac{d}{dt}$$

8. Jul 20, 2009

### jeff1evesque

So anything involving derivatives (rates), in particular capacitors only have imaginary components? And anything that doesn't involve derivatives, such as resistors- like lightbulbs or whatever- only have real components? If this is true, do you mind explaining why that is so?

Thanks so much,

JL

Last edited: Jul 20, 2009
9. Jul 20, 2009

### CEL

Not only elements involving derivatives. Elements involving integrals too. In the case of electrical elements, these involve capacitors and inductors.
If you apply the Laplace transform to an integro-differential equation, the derivatives become s and the integrals, 1/s, where $$s = \sigma + j\omega$$ is the Laplace variable.
The real term $$\sigma$$ corresponds to the transient response, while the imaginary term $$j\omega$$ corresponds to the steady state response.
If you are interested only in the steady state, you replace s by $$j\omega$$ .