Blog #53 - Impedance Problem

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In this problem we are too calculate the current in this circuit for that we need to know the total impedance of the circuit. The angular frequency is given in hertz or cycles per second but we needed in radians per second so converting it…the angular frequency ω = 2π times the frequency in Hz f or…(2)(3.1416)(60)… which is 377 rad/sec.

∴ XL is equal to ωL…is (377)(0.5) = 188.5 Ω…and XC = minus 1/(ωC)… plugging in our actual values - 1/[377)(26.5)x10-6] which is = - 100 Ω.

Let’s let the total reactive impedance of the circuit be XT which is XL plus XE…

Then Z = R + j(XL  - XC) = R + jXT 100 - j88.5 in the rectangular coordinate format… or in polar coordinate format…Z = 133.5 @41.5° Ω

The impedance triangle here illustrates the results of the preceding solution.

From Ohm’s law for AC circuits, the current I can be described by the applied voltage E divided by the circuit impedance Z all of these of course are phasors…

I am letting the source voltage E be reference here so 120 Volts at 0 degrees looks like this…and the impedance Z being equal to 133.5∠41.5 deg. Therefore the current I = 120∠0 deg/133.5∠41.5 deg…or  .899∠-41.5 Volts…which looks like the above…

Which means the current will be lagging the reference voltage by 41.5 deg.

Note: in my diagram, in order to fit the phasors comfortably on the slide, the magnitude of the current is exaggerated compared to the voltage, however the angles are correct.

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