voltage The circuit in Fig. contains a 3.60-ml 25.0 V. Find the rms current in the circuit when the generator frequency is (a) 1.00 x 10² 10³ Hz. Hz and (b) 5.00 x
voltage The circuit in Fig. contains a 3.60-ml 25.0 V. Find the rms current in the circuit when the generator frequency is (a) 1.00 x 10² 10³ Hz. Hz and (b) 5.00 x
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![### Analysis of an Inductive AC Circuit
The circuit in the figure contains a 3.60-mH inductor. The RMS (Root Mean Square) voltage of the generator is 25.0 V. Determine the RMS current in the circuit when the generator frequency is:
a) \( 1.00 \times 10^2 \) Hz
b) \( 5.00 \times 10^3 \) Hz
#### Diagram Explanation
- **Components:**
- **Inductor (L):** Represented by a coil in the diagram.
- **AC Voltage Source (\( V_0 \sin 2 \pi ft \)):** The AC voltage source is shown by a sine wave symbol with voltage \( V_0 \) and frequency \( f \).
The given inductor is connected in a simple series circuit with an AC voltage source.
#### Circuit Parameters:
- Inductance (\( L \)) = 3.60 mH (millihenries)
- RMS Voltage (\( V_{rms} \)) = 25.0 V
- Frequency (\( f \)) for two scenarios:
- (a) \( f = 1.00 \times 10^2 \) Hz (100 Hz)
- (b) \( f = 5.00 \times 10^3 \) Hz (5000 Hz)
#### Calculation of RMS Current
To find the RMS current, we use the following steps:
1. **Inductive Reactance ( \( X_L \) ) Calculation:**
The inductive reactance is given by:
\[
X_L = 2 \pi f L
\]
- For \( f = 1.00 \times 10^2 \) Hz:
\[
X_L = 2 \pi (1.00 \times 10^2) \times 3.60 \times 10^{-3}
\]
- For \( f = 5.00 \times 10^3 \) Hz:
\[
X_L = 2 \pi (5.00 \times 10^3) \times 3.60 \times 10^{-3}
\]
2. **RMS Current (\( I_{rms} \)) Calculation:**
The RMS current is given by Ohm's Law for AC circuits:
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Transcribed Image Text:### Analysis of an Inductive AC Circuit
The circuit in the figure contains a 3.60-mH inductor. The RMS (Root Mean Square) voltage of the generator is 25.0 V. Determine the RMS current in the circuit when the generator frequency is:
a) \( 1.00 \times 10^2 \) Hz
b) \( 5.00 \times 10^3 \) Hz
#### Diagram Explanation
- **Components:**
- **Inductor (L):** Represented by a coil in the diagram.
- **AC Voltage Source (\( V_0 \sin 2 \pi ft \)):** The AC voltage source is shown by a sine wave symbol with voltage \( V_0 \) and frequency \( f \).
The given inductor is connected in a simple series circuit with an AC voltage source.
#### Circuit Parameters:
- Inductance (\( L \)) = 3.60 mH (millihenries)
- RMS Voltage (\( V_{rms} \)) = 25.0 V
- Frequency (\( f \)) for two scenarios:
- (a) \( f = 1.00 \times 10^2 \) Hz (100 Hz)
- (b) \( f = 5.00 \times 10^3 \) Hz (5000 Hz)
#### Calculation of RMS Current
To find the RMS current, we use the following steps:
1. **Inductive Reactance ( \( X_L \) ) Calculation:**
The inductive reactance is given by:
\[
X_L = 2 \pi f L
\]
- For \( f = 1.00 \times 10^2 \) Hz:
\[
X_L = 2 \pi (1.00 \times 10^2) \times 3.60 \times 10^{-3}
\]
- For \( f = 5.00 \times 10^3 \) Hz:
\[
X_L = 2 \pi (5.00 \times 10^3) \times 3.60 \times 10^{-3}
\]
2. **RMS Current (\( I_{rms} \)) Calculation:**
The RMS current is given by Ohm's Law for AC circuits:
\[
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