inductor are connected in series. The resistor has a resistance of 50.0 N, the capacitor has a capacitance of 2.50 µF and the inductor has an inductance of 2.00 mH (milli Henry's). The circuit is subject to an AC signal with a frequency of 1000 hz with an RMS Voltage of 10.0 V.
inductor are connected in series. The resistor has a resistance of 50.0 N, the capacitor has a capacitance of 2.50 µF and the inductor has an inductance of 2.00 mH (milli Henry's). The circuit is subject to an AC signal with a frequency of 1000 hz with an RMS Voltage of 10.0 V.
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Chapter1: Units, Trigonometry. And Vectors
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![**AC Circuits, Reactance, and Resonance**
A resistor, capacitor, and inductor are connected in series. The resistor has a resistance of 50.0 Ω, the capacitor has a capacitance of 2.50 μF, and the inductor has an inductance of 2.00 mH. The circuit is subject to an AC signal with a frequency of 1000 Hz and an RMS Voltage of 10.0 V.
**a. Angular Frequency Calculation**
- **Question**: What is the angular frequency ω associated with the AC signal?
- **Solution**: The angular frequency is calculated as:
\[
\text{angular frequency} = (6.25)(.2)^2 = 1.200 f
\]
**b. Impedance Calculation**
- **Question**: What is the impedance of the circuit?
- **Solution**: The impedance is given as:
\[
50 \, \text{ohms}
\]
**c. Resonant Frequency Calculation**
- **Question**: What is the resonant frequency of the circuit?
- **Solution**: The resonant frequency is:
\[
2000 \, \text{Hz}
\]
**d. RMS Current Calculation**
- **Question**: What RMS current flows in the circuit?
- **Solution**: The relationship is given by:
\[
V_{RMS} = I_{RMS} Z
\]
For a 10.0 V RMS and circuit impedance at 1600 Hz:
\[
\text{RMS current} = 0 \, A
\]
**Notes**:
- The calculations involve typical formulas used in AC circuit analysis: angular frequency \( \omega = 2\pi f \), impedance in an RLC series circuit \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \).
- The solution for part (d) indicates that at certain frequencies, the voltage might result in no observable RMS current, suggesting potential cancellations at specific frequencies, often resonant or due to rounding or miscalculations in context.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa442f039-c23f-4767-9723-571e7d7e1a22%2F71ede8a3-1f45-4bb4-9010-5911f2b310e2%2Fmtdkgbe.jpeg&w=3840&q=75)
Transcribed Image Text:**AC Circuits, Reactance, and Resonance**
A resistor, capacitor, and inductor are connected in series. The resistor has a resistance of 50.0 Ω, the capacitor has a capacitance of 2.50 μF, and the inductor has an inductance of 2.00 mH. The circuit is subject to an AC signal with a frequency of 1000 Hz and an RMS Voltage of 10.0 V.
**a. Angular Frequency Calculation**
- **Question**: What is the angular frequency ω associated with the AC signal?
- **Solution**: The angular frequency is calculated as:
\[
\text{angular frequency} = (6.25)(.2)^2 = 1.200 f
\]
**b. Impedance Calculation**
- **Question**: What is the impedance of the circuit?
- **Solution**: The impedance is given as:
\[
50 \, \text{ohms}
\]
**c. Resonant Frequency Calculation**
- **Question**: What is the resonant frequency of the circuit?
- **Solution**: The resonant frequency is:
\[
2000 \, \text{Hz}
\]
**d. RMS Current Calculation**
- **Question**: What RMS current flows in the circuit?
- **Solution**: The relationship is given by:
\[
V_{RMS} = I_{RMS} Z
\]
For a 10.0 V RMS and circuit impedance at 1600 Hz:
\[
\text{RMS current} = 0 \, A
\]
**Notes**:
- The calculations involve typical formulas used in AC circuit analysis: angular frequency \( \omega = 2\pi f \), impedance in an RLC series circuit \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \).
- The solution for part (d) indicates that at certain frequencies, the voltage might result in no observable RMS current, suggesting potential cancellations at specific frequencies, often resonant or due to rounding or miscalculations in context.
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