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Does anyone know how to do this

A circuit having a resistance of 12 ohm, an indutance of 96mH and a capacitance of 52mF in series, is connected across a 220V, 50Hz supply. Calculate the following:

Total Impedance

current flowing through the circuit

the voltage across each component

the phase difference between the current and the supply voltage

The power factor of the circuit

2007-06-28 18:48:21 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

The total impedance is Z(w) = R + jwL + 1/jwC

w = angular frequency = 2*π*f
j = √-1

The total current is V / Z

The voltage across each component is the impedance of that component times the current through it. The current is the same for all, inductor impedance is jwL, capacitive impedance is 1/jwC.

When you calculate the current, it will be in the form Ir + j*Ii; the phase is arctan(Ii/Ir).

The power factor is cos(ø), where ø is the phase.

All calcs must be carried out with complex numbers.

I assumed mF = microFarad, while mH is milliHenry, which are common units for these components. If this is not right (then they are either both milli or both micro), you do the computation for the proper values.

Z = 12 - 31.05j ohms

I = 0.54 + 1.4j amps

Vr = 6.5 + 16.8j volts

VL = -42.3 + 16.3j volts

VC = 85.75 - 33.1j volts

ø = 68.9º

PF = 0.36

Note: in calculating with complex numbers, adding is easier in a = bj form, but multiplying and dividing is easier in polar form. Convert according to the following formula:

a + bj = (√[a^2 + b^2]) | arcan(b/a) = A | ø

To multiply A1 | ø1 by A2 | ø2, A1*A2 | ø1 + ø2
To divide A1 | ø1 by A2 | ø2, A1/A2 / ø1 - ø2

To get back to rectangular notation, use a = A*cosø, b = A*sinø

2007-06-28 19:39:12 · answer #1 · answered by gp4rts 7 · 0 0

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