Calculating Impedance and Power in AC Circuits

In summary, we have a circuit with an impedance of 8 + j7 Ω connected in parallel with another impedance of 5 + j6 Ω, which is then connected in series with a resistance of 5 Ω and a capacitive reactance of 7 Ω. The circuit is connected to a 150 volt, 50 hertz supply. From this, we can calculate the supply current, the circuit power factor, and the potential drop across each circuit. To find the potential drop, we can use Ohm's law and the known values of current and impedance for each series circuit block.
  • #1
Michael V
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0

Homework Statement



An impedance 8 + j7 Ω is connected in parallel with another impedance of 5 + j6 Ω. this circuit is then connected in series with another impedance, comprising a resistance of 5 Ω in series with a capacitive reactance of 7 Ω. The complete circuit is then connected to 150 volt, 50 hertz supply.

Calculate the following:

1. The supply current
2. The circuit power factor
3. The potential drop across each circuit.

Homework Equations



ZT = (Z1 × Z2)/(Z1 + Z2)

The Attempt at a Solution



I attached my attempt. Having trouble with question 3. Potential drop?
 

Attachments

  • attempted answer.pdf
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Last edited:
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  • #2
You should keep a few more digits in your intermediate results; roundoff error is creeping into your significant figures as you go further into the calculations.

For potential drop, you have the current and the impedance of each of the series circuit 'blocks' (your Zp and Z3). Apply Ohm's law.
 
  • #3
OK, will do. So the potential drop is the same as voltage drop or potential difference? I attached my last answer, does it look ok? Thanks for help.
 

Attachments

  • answer to 3.pdf
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  • #4
Michael V said:
OK, will do. So the potential drop is the same as voltage drop or potential difference? I attached my last answer, does it look ok? Thanks for help.

Yup. And you're method's fine.

It's not clear to me from the problem statement whether you need to find the potentials across the individual components of the series circuit z3, too; While parallel components always share the same potential difference, series ones share the same current but have different potential differences.
 
  • #5
I see. Thanks for your help.
 

Related to Calculating Impedance and Power in AC Circuits

1. What is impedance in polar form?

Impedance in polar form is a way of representing the complex impedance of a circuit using polar coordinates. It consists of a magnitude (or amplitude) and a phase angle, which together determine the overall impedance of the circuit.

2. How is impedance in polar form different from impedance in rectangular form?

Impedance in polar form and impedance in rectangular form are two different ways of representing the same complex impedance. While polar form uses magnitude and phase angle, rectangular form uses real and imaginary components. Both forms can be converted to each other using simple mathematical equations.

3. What is the significance of the phase angle in impedance in polar form?

The phase angle in impedance in polar form represents the phase shift between the voltage and current in a circuit. It determines the amount by which the current lags or leads the voltage, and is important in analyzing the behavior and performance of electrical systems.

4. How is impedance in polar form used in circuit analysis?

Impedance in polar form is used in circuit analysis to simplify complex calculations and make it easier to understand the behavior of a circuit. It allows us to combine impedances using simple addition and subtraction, and also makes it easier to visualize the effects of different components on the overall impedance.

5. Can impedance in polar form be used for all types of circuits?

Yes, impedance in polar form can be used for all types of circuits, including AC circuits and filters. It is especially useful for circuits with multiple components, as it allows us to easily determine the overall impedance and analyze the behavior of the circuit.

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