created at August 28, 2021

# Parallel RC Circuit

A parallel RC circuit contains a resistor and a capacitor connected in parallel:

$u = U_{m}\cdot sin(\omega\cdot t)$

The current IR flowing through the resistor has the same phase angle as the voltage U:

$u = U_{m}\cdot sin(\omega\cdot t)$

The capacitor's current IC outstrips the voltage by phase angle π/2:

$i_{C} = \frac{U_{m}}{X_{C}} sin \left ( \omega\cdot t + \frac{\pi}{2} \right ); \quad X_{C} = \frac{1}{\omega\cdot C}$

The total current I is a sum of IR and IC (the first Kirchhoff's law). The vector diagram for all currents (current triangle):

hence:

$I = \sqrt{I_{R}^2 + I_{C}^2}$

The current I outstrips the voltage by phase angle φ:

$cos\phi = \frac{I_{R}}{I}$

The angle φ is being counted starting from the current I vector, hence in our case it is negative as it was for serial RC circuit.

After dividing the current triangle by the voltage U, we are getting the conductivity triangle, but if we multiply by U - we get the power triangle:

then:

$Y = \sqrt{G^2 + B_{C}^2}$

where G = IA/U - conductivity of resistive branch, BC = IC/U - conductivity of capacitive branch, Y = I/U - the total conductivity.

The impedance Z (full resistance) is a reverse value to conductivity:

$Z = \frac{1}{\sqrt{G^2 + B_C^2}} = \frac{1}{\sqrt{\frac{1}{R^2} + \frac{1}{\omega^2\cdot C^2}}}$

The total power:

$S = \sqrt{P^2 + Q_C^2}$

where P = IA·U, QC = IC·U.