Solving a Complex Series Circuit Problem: Help Needed!

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SUMMARY

The discussion focuses on solving a complex series circuit problem involving a 1μF capacitor, a 50Ω resistor, and a 49mH inductor with an AC current of 0.03A at 500Hz. The peak potential difference (p.d.) across the combination is calculated using the total impedance formula, resulting in a value of 2.16V. This value is derived from the impedance of the circuit, which is 50.85Ω, and the peak current calculated as 0.0424A. The individual peak voltages across the capacitor and inductor cannot be simply summed due to their phase differences.

PREREQUISITES
  • Understanding of AC circuit analysis
  • Familiarity with impedance calculations
  • Knowledge of phasors and phase relationships in circuits
  • Proficiency in using Ohm's Law in AC circuits
NEXT STEPS
  • Study the concept of impedance in AC circuits, focusing on the formula Z = √(R² + (XL - XC)²)
  • Learn about phasor diagrams and their application in analyzing AC circuits
  • Explore the calculation of inductive reactance (XL) and capacitive reactance (XC)
  • Investigate the relationship between RMS and peak values in AC current
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing AC circuits will benefit from this discussion.

Clari
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Hi there! I have some problems in this topic...real difficult to me. ^-^

A 1(miu)F capacitor is connected in series with a coil of R=50(ohm) and L=49mH and an a.c. ammeter. A current of r.m.s. value 0.03A and f=500Hz is passes through the combination. What is the peak p.d. across the combination?

That question is in part c, that means i have solved the peak p.d. across the capacitor(which is=13.5V) and that across the coil(which is=6.8V). well, i know the answer should not be (13.5V+6.8V), but then i don't know how... :frown:

please help me.
 
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Clari said:
Hi there! I have some problems in this topic...real difficult to me. ^-^

A 1(miu)F capacitor is connected in series with a coil of R=50(ohm) and L=49mH and an a.c. ammeter. A current of r.m.s. value 0.03A and f=500Hz is passes through the combination. What is the peak p.d. across the combination?

That question is in part c, that means i have solved the peak p.d. across the capacitor(which is=13.5V) and that across the coil(which is=6.8V). well, i know the answer should not be (13.5V+6.8V), but then i don't know how... :frown:

please help me.

The approach will depend on what methods you have learned. I'm guessing that you have learned about phasors, which treat the potential differences across elements as rotating vectors with some angle between them. That angle represents the phase difference between the potentials for the different elements. The p.d. for the resistor will be in phase with the current. The p.d. for the coil will lead the current by 90 degrees and the p.d. for the capacitor will lag the current by 90 degrees. Have you heard of ELI the ICE man?

Another approach is to find the equivalent impedence of the circuit, for which you should have a formula involving a square root of some combination of circuit element parameters and the frequency. The product of the equivalent impedence times the current will give the p.d. of the combination.
 


Hi there! It looks like you're having trouble solving a complex series circuit problem. Don't worry, I'm here to help! Let's break down the problem and tackle it step by step.

First, let's calculate the total impedance of the circuit. We know that impedance is the combination of resistance and reactance in a circuit. In this case, we have a capacitor and a coil, which have different reactances.

To find the total impedance, we use the formula Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

So, Z = √(50^2 + (2πfL - 1/(2πfC))^2), where f is the frequency, L is the inductance, and C is the capacitance.

Plugging in the values given in the problem, we get Z = √(50^2 + (2π(500)(0.049) - 1/(2π(500)(1x10^-6)))^2 = 50.85 ohms.

Next, we can use Ohm's law (V = IZ) to find the peak voltage. We know the r.m.s. value of the current (0.03A), so we can use that to find the peak current (I = √2 x 0.03 = 0.0424A).

Now, plugging in the values, we get V = (0.0424)(50.85) = 2.16V.

Therefore, the peak p.d. across the combination of the capacitor and coil is 2.16V. This is not the sum of the individual peak voltages across the capacitor and coil because they are not in phase with each other and their voltages do not add up directly.

I hope this helps you understand how to solve the problem! If you have any further questions, feel free to ask. Good luck!
 

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