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Posted by George on October 13, 2011 at 00:32:22:

In Reply to: electircal math question posted by shaun on February 02, 2011 at 19:51:26:

: Been using the Capacitance Reactance formula (impedence) to try and solve this question. I have the answer (8 micro farads) but am having trouble solving. Any help would be appreciated, thanks

: Question:

: A 220V, 60 Hz ac circiut draws .857 amps. It contains a 10 ohm resistor, a 200 milli-henry induction coil and a condenser all connected in series. Find the conductance of the condenser in farads.

Ok, first, a capacitor's capacity (capacitance) is measured in farads, not the conductance. If you're looking for conductance (1/impedance), you need to calculate the capacitive reactance.

Impedance (Z, ohms) in a combined Capacitive and Reactive circuit = SQRT{R^2 + [X(LC]^2}.

X(L) = 2 (pi) (F) (L) and

X(C) = 2 (pi) (F) (C), and

X(LC) = 2 (pi) (F) (L) - 1/2 2 (pi) (F) (C).

Since the total amp draw for the circuit is .857 Amps and the applied voltage is 220 Volts, the circuit impedance, Z = 220 / .857 = 256.709 Ohms, of which 10 Ohms are from the resistor, leaving 256.709-10=246.709 Ohms consisting of combined Capacitive and Inductive Reactances. The Inductive Reactance, X(L)=(2 x 3.1416 x 60 x .2) =

75.4 Ohms, leaving 246.709 - 75.4 = 171.309 (say 171.3) Ohms offered by the capacitor. Since Capacitive Reactance, X(C) = 1/2 (pi) (F) (C), C=1/X(C) (2) (pi) (F), or 1/171.3 X 2 X 3.1416 x 60 or .15 uF, or 150 pF. Show me where I'm wrong, please somebody...