Steady State Circuit Analysis with Phasors

AI Thread Summary
The discussion focuses on solving a steady-state circuit analysis problem using phasors, specifically calculating current and equivalent impedance. The initial impedance calculation was incorrect, leading to confusion about the correct values. Participants clarified that the final impedance should be approximately 123.875 Ω at an angle of 7.532°, with no need to apply RMS values to impedance calculations. It was emphasized that the input voltage can remain in peak form for calculations, and care should be taken when determining the phase of the current, as it is the negative of the impedance's angle. Accurate calculations and understanding of phasor relationships are crucial for solving such circuit problems effectively.
alexmath
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Homework Statement



Hello everyone! I have the following circuit to solve, and my result is a bit wrong... can you tell me please where's the mistake?

E=10sin(1000t)

Find the current delivered to the circuit. Find the equivalent impedance of the circuit. Find the equation of the current and voltage drop for each element of the circuit.

Homework Equations



I tried first to solve for the equivalent impedance.

Vrms=10/√2
Xc= -j * 1/wc
Xl = jwl
w=1000 rad/s

The Attempt at a Solution



R1=100Ω, R2=20Ω, R3=50Ω, Xc=-100j, Xl=40j
therefore equivalent impedance is: 100+40j+ ((1/(50-100j))+0.05)^(-1)=40j+100, in polar form: 107.7 < 21.8, so the impedance=107.7/√2(1000t+ 21.8°)Ω.
Is that correct?
 

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Check your math on the impedance calculation. Your expression looks okay but your result does not.
 
checked it on wolframalpha... what's wrong? :(
 
alexmath said:
checked it on wolframalpha... what's wrong? :(

It's not the same as the result I get. Maybe break down your calculations and show some intermediate results? Start with the parallel combination of the 20 Ω resistor with the capacitor and 50 Ω resistor. What impedance do you get for those components ?
 
only two first components added up give 40j+100 ( i did not even check the calculation) , wolframalpha was wrong haha... the correct answer is:

123.47<17.53 right?

so the final impedance is 123.47 over sqrt 2 or not here? (1000t + 17.53 degrees converted in radians)
 
alexmath said:
only two first components added up give 40j+100 ( i did not even check the calculation) , wolframalpha was wrong haha... the correct answer is:

123.47<17.53 right?
Close enough! I'm seeing (extra digits kept for intermediate value to prevent roundoff/truncation errors creeping into future calculations) Z = 123.875 Ω ∠7.532° .

so the final impedance is 123.47 over sqrt 2 or not here? (1000t + 17.53 degrees converted in radians)

No root 2 involved in impedance ... RMS values apply to voltages and currents. Usually it's best just to convert the input voltage to an RMS value right at the start and then everything will take care of itself from then on.

In this problem, since you aren't calculating any powers (for which RMS values are key), you could get away with leaving the input voltage as 10 V (peak), do the calculations for the circuit, and write the time domain results as peak voltages.

Take care when you calculate the phase of the current! Your impedance is in the denominator of I = E/Z, so the angle of the current will be the negative of the angle of the impedance (assuming that the voltage E is the reference phasor with angle zero).
 
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