Engineering Solve Series RC Circuit for Capacitance & Currents

[fonz]

Homework Statement

Three lamps are rated 110 V, 60 W. They are connected in parallel and a capacitor is connected in series with the group. The circuit is then connected to a 230 V 50 Hz power supply. Detremine:

a) The capacitance which is required to provide the correct voltage across the lamps.
b) The active and reactive currents.

Homework Equations

$P = 60 W, U_S = 230 V, U_R = 110 V, f = 50 Hz$

$C = \frac {1} {2 \pi f X_C}$

The Attempt at a Solution

Part a)

$U_C = \sqrt {230^2 - 110^2} = 202 V$

$I = \frac {3P} {U_R} = \frac {3*60} {110} = 1.63 A$

$X_C = \frac {U_C} {I} = \frac {202} {1.63} = 124 \Omega$

$C = \frac {1} {2 \pi f X_C} = \frac {1} {2*110*50 \pi} = 25.7 \mu F$

Part b)

$\phi = \cos^{-1} (\frac {U_R} {U_S}) = \cos^{-1} (\frac {110} {230}) = 61.4^{\circ}$

$I_C = I \sin (\phi) = 1.63 \sin (61.4) = 1.42 A$

$I_R = I \cos (\phi) = 1.63 \cos (61.4) = 0.79 A$

I'm pretty confident this is correct but really my question is: is there a way of answering part b without working out the phase difference?

 

[gneill]

I'm pretty confident this is correct but really my question is: is there a way of answering part b without working out the phase difference?

You could do the computation using impedance (complex arithmetic). Then it's just Ohm's law to find the circuit's current.

 

[fonz]

You could do the computation using impedance (complex arithmetic). Then it's just Ohm's law to find the circuit's current.

Thanks for the help.