- #1
jeff1evesque
- 312
- 0
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:
[tex]I = GV + C\frac{dV}{dt} = GV + j\omega CV[/tex] (#1)
I know [tex]I = \frac{V}{R}, C\frac{dV}{dt}[/tex] 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 [tex]GV + C\frac{dV}{dt}[/tex]
thank you,
JL
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:
[tex]I = GV + C\frac{dV}{dt} = GV + j\omega CV[/tex] (#1)
I know [tex]I = \frac{V}{R}, C\frac{dV}{dt}[/tex] 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 [tex]GV + C\frac{dV}{dt}[/tex]
thank you,
JL
Last edited: