Time Averages and Phasors in Electrical Circuits

[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

 

[CEL]

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

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).

 

[jeff1evesque]

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).

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

 

[CEL]

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

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

 

[jeff1evesque]

I'm guessing that's 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

 

[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 $$C\frac{d(Vcos(\omega t))}{dt} = \frac{Vsin(\omega t + \theta)}{\frac{1}{\omega C}}}.$$

 

[CEL]

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 $$C\frac{d(Vcos(\omega t))}{dt} = \frac{Vsin(\omega t + \theta)}{\frac{1}{\omega C}}}.$$

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

 

[jeff1evesque]

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

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

 

[CEL]

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

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 $$j\omega$$ .