A series RLC circuit contains a 4-kO resistor, an inductor with an inductive reactance (X,) of 3.5 kN, and a capacitor with a capacitive reactance (Xc) of 2.4 kN. A 120-Vac, 60-Hz power source is connected to the circuit. How much voltage is dropped across the inductor?
A series RLC circuit contains a 4-kO resistor, an inductor with an inductive reactance (X,) of 3.5 kN, and a capacitor with a capacitive reactance (Xc) of 2.4 kN. A 120-Vac, 60-Hz power source is connected to the circuit. How much voltage is dropped across the inductor?
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![**RLC Circuit Voltage Drop Calculation**
---
In this example, we analyze a series RLC circuit consisting of different electrical components and determine the voltage drop across the inductor.
**Components and Given Values:**
- **Resistor (R):** 4 kilo-ohms (kΩ).
- **Inductor with Inductive Reactance (XL):** 3.5 kilo-ohms (kΩ).
- **Capacitor with Capacitive Reactance (XC):** 2.4 kilo-ohms (kΩ).
- **Power Source:** 120 volts AC (Vac), 60 hertz (Hz).
**Problem Statement:**
A series RLC circuit contains a 4-kΩ resistor, an inductor with an inductive reactance (XL) of 3.5 kΩ, and a capacitor with a capacitive reactance (XC) of 2.4 kΩ. A 120-Vac, 60-Hz power source is connected to the circuit. How much voltage is dropped across the inductor?
**Solution:**
1. **Calculate the Impedance (Z) of the Circuit:**
- The impedance, \( Z \), in a series RLC circuit is determined as:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
- Substituting the values:
\[
Z = \sqrt{(4000)^2 + (3500 - 2400)^2} = \sqrt{(4000)^2 + (1100)^2} = \sqrt{16000000 + 1210000} = \sqrt{17210000} \approx 4148 \text{ ohms}
\]
2. **Calculate the Current (I) in the Circuit:**
- The current, \( I \), through the series circuit is determined by Ohm's law using the total voltage (V) and impedance (Z):
\[
I = \frac{V}{Z} = \frac{120}{4148} \approx 0.0289 \text{ amps (A)}
\]
3. **Calculate the Voltage Drop across the Inductor (V_L):**
- The voltage drop across the inductor is given by:
\[
V_L = I \times X_L
\]
- Substituting the values:
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Transcribed Image Text:**RLC Circuit Voltage Drop Calculation**
---
In this example, we analyze a series RLC circuit consisting of different electrical components and determine the voltage drop across the inductor.
**Components and Given Values:**
- **Resistor (R):** 4 kilo-ohms (kΩ).
- **Inductor with Inductive Reactance (XL):** 3.5 kilo-ohms (kΩ).
- **Capacitor with Capacitive Reactance (XC):** 2.4 kilo-ohms (kΩ).
- **Power Source:** 120 volts AC (Vac), 60 hertz (Hz).
**Problem Statement:**
A series RLC circuit contains a 4-kΩ resistor, an inductor with an inductive reactance (XL) of 3.5 kΩ, and a capacitor with a capacitive reactance (XC) of 2.4 kΩ. A 120-Vac, 60-Hz power source is connected to the circuit. How much voltage is dropped across the inductor?
**Solution:**
1. **Calculate the Impedance (Z) of the Circuit:**
- The impedance, \( Z \), in a series RLC circuit is determined as:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
- Substituting the values:
\[
Z = \sqrt{(4000)^2 + (3500 - 2400)^2} = \sqrt{(4000)^2 + (1100)^2} = \sqrt{16000000 + 1210000} = \sqrt{17210000} \approx 4148 \text{ ohms}
\]
2. **Calculate the Current (I) in the Circuit:**
- The current, \( I \), through the series circuit is determined by Ohm's law using the total voltage (V) and impedance (Z):
\[
I = \frac{V}{Z} = \frac{120}{4148} \approx 0.0289 \text{ amps (A)}
\]
3. **Calculate the Voltage Drop across the Inductor (V_L):**
- The voltage drop across the inductor is given by:
\[
V_L = I \times X_L
\]
- Substituting the values:
\
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