How to Calculate Current in a Parallel RLC Circuit at 1kHz?

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To calculate the current in a parallel RLC circuit with a 50 mH inductor, a 10 µF capacitor, and a 200-ohm resistor at 1 kHz, it is essential to recognize that the impedances are combined in parallel, not series. The correct approach involves calculating the individual impedances for the resistor, inductor, and capacitor, and then using the formula for parallel impedance. The net impedance is determined using the formula Z = 1 / (1/Z_R + 1/Z_C + 1/Z_L). After finding the total impedance, the current can be calculated using I = V/Z. This method yields a current of approximately 0.59A, aligning with the expected result.
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Homework Statement



A 50 mH inductor, a 10 microF capacitor and a 200 ohm resistor are connected in parallel. A voltage of 10 v at 1kHz is applied. what is the magnitude of current that flows?

Homework Equations


The Attempt at a Solution



I recognised this would be a problem where i would need to use V=IZ

The formula I'm familiar with is

V = I (R + i ωL -i/(ωc)

Calculating ω = 2pi * f i get ω= 2pi * 10^3

To calculate the magnitude of z i added the entire thing in quadrature so

|Z| = (R^2 + (ωL -1/(ωc))^2)^0.5

which came out as 359.095

from this i subbed it in and got a value for current as 0.0278A
The answers given say it should be 0.59A
 
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You're thinking along the right lines, but in this case all the impedances are in parallel, not series. Parallel impedances combine just like parallel resistors, only you're dealing with complex numbers.

Start by determining numerical values for each of the impedances involved:

ZR = ?
ZL = ?
ZC = ?

Then calculate the net impedance as their parallel combination:

Z = \frac{1}{\frac{1}{Z_R} + \frac{1}{Z_C} + \frac{1}{Z_L}}

Then do your I = V/Z stuff.
 
You found the current for the situation in which R, C, & L are in series.

In parallel, \frac{1}{|Z|}=\sqrt{\frac{1}{R^2}+\left(\omega C-\frac{1}{\omega L}\right)^2}.

The voltage across all the devices is in phase. The currents are out of phase .
 
Tried it, and worked! thanks!
 
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