# Steady State Circuit Analysis with Phasors

1. Apr 20, 2014

### alexmath

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

I tried first to solve for the equivalent impedance.

Vrms=10/√2
Xc= -j * 1/wc
Xl = jwl

3. 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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Last edited: Apr 20, 2014
2. Apr 20, 2014

### Staff: Mentor

3. Apr 20, 2014

### alexmath

checked it on wolframalpha... what's wrong? :(

4. Apr 20, 2014

### Staff: Mentor

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 ?

5. Apr 20, 2014

### alexmath

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)

6. Apr 20, 2014

### Staff: Mentor

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° .

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).