assume that tthe circuit shown in figure 21-1 is connected to a 480 v 60 hz line. the capcitor has a capacitance of 165.782 uf and the resistor has a resistance of 12 ohmas find the missing valuses.

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assume that tthe circuit shown in figure 21-1 is connected to a 480 v 60 hz line. the capcitor has a capacitance of 165.782 uf and the resistor has a resistance of 12 ohmas find the missing valuses.

The diagram represents an electrical circuit with both resistive and capacitive components. Here is a detailed explanation of the circuit components:

1. **Voltage Source**: The circuit is powered by an AC voltage source labeled as "ET 240 V," indicating that it provides 240 volts.

2. **Resistor**: 
   - Labeled "R."
   - Has a resistance value of 12 ohms (Ω).
   - Associated parameters include:
     - \( IR \): Current through the resistor.
     - \( ER \): Voltage across the resistor.
     - \( P \): Power dissipated by the resistor.

3. **Capacitor**:
   - Labeled "C."
   - Has a capacitive reactance of \( X_C = 16 Ω \).
   - Associated parameters include:
     - \( I_C \): Current through the capacitor.
     - \( E_C \): Voltage across the capacitor.
     - \( \text{VARSc} \): Reactive power in volt-amperes reactive (VARs).

4. **Measurement Point**:
   - There is a measurement point, possibly representing an ammeter or voltmeter (the symbol near "VA").
   - The following parameters might be measured:
     - \( I_T \): Total current in the circuit.
     - Z: Impedance of the circuit.
     - \( \text{VA} \): Apparent power in volt-amperes (VA).
     - \( \text{PF} \): Power factor, with an angle \( θ \).

**Figure Reference**: 
- The diagram is labeled as "FIGURE 21-1" in an educational context, likely part of a textbook or a course material discussing AC circuits and the interaction between resistive and capacitive components.

This representation can be used to study the behavior of AC circuits, power calculation, impedance, and phase relationships between current and voltage.
Transcribed Image Text:The diagram represents an electrical circuit with both resistive and capacitive components. Here is a detailed explanation of the circuit components: 1. **Voltage Source**: The circuit is powered by an AC voltage source labeled as "ET 240 V," indicating that it provides 240 volts. 2. **Resistor**: - Labeled "R." - Has a resistance value of 12 ohms (Ω). - Associated parameters include: - \( IR \): Current through the resistor. - \( ER \): Voltage across the resistor. - \( P \): Power dissipated by the resistor. 3. **Capacitor**: - Labeled "C." - Has a capacitive reactance of \( X_C = 16 Ω \). - Associated parameters include: - \( I_C \): Current through the capacitor. - \( E_C \): Voltage across the capacitor. - \( \text{VARSc} \): Reactive power in volt-amperes reactive (VARs). 4. **Measurement Point**: - There is a measurement point, possibly representing an ammeter or voltmeter (the symbol near "VA"). - The following parameters might be measured: - \( I_T \): Total current in the circuit. - Z: Impedance of the circuit. - \( \text{VA} \): Apparent power in volt-amperes (VA). - \( \text{PF} \): Power factor, with an angle \( θ \). **Figure Reference**: - The diagram is labeled as "FIGURE 21-1" in an educational context, likely part of a textbook or a course material discussing AC circuits and the interaction between resistive and capacitive components. This representation can be used to study the behavior of AC circuits, power calculation, impedance, and phase relationships between current and voltage.
**Problem Statement:**

1. Assume that the circuit shown in Figure 21-1 is connected to a 480-V, 60-Hz line. The capacitor has a capacitance of 165.782 µF, and the resistor has a resistance of 12 Ω. Find the missing values.

**Given Data:**
- Voltage, \( E = 480 \) V
- Resistance, \( R = 12 \) Ω
- Capacitance, \( C = 165.782 \) µF

**To Find:**

- \( Z \) (Impedance)
- \( I_T \) (Total current)
- \( E_R \)
- \( I_R \)
- \( E_C \)
- \( I_C \)
- \( X_C \) (Capacitive reactance)
- \( VARS \)
- \( VA \)
- \( PF \) (Power factor)
- \( P \) (Real power)
- \( \theta \) (Phase angle)

**Explanation:**
The setup involves a resistor and capacitor connected to an AC source. The information provided suggests a need to calculate the collective properties of impedance in the circuit, analyze individual components, and determine both the real and reactive power.

**Calculation Guidance:**
1. Use the formulas for capacitive reactance and impedance:
   \[
   X_C = \frac{1}{2 \pi f C}
   \]
   \[
   Z = \sqrt{R^2 + X_C^2}
   \]

2. Find the currents and voltages across the resistor and capacitor using Ohm's Law:
   \[
   I_R = \frac{E_R}{R}
   \]
   \[
   I_C = \frac{E_C}{X_C}
   \]

3. Calculate power values:
   \[
   P = VI \cos \theta
   \]
   \[
   VARS = VI \sin \theta
   \]

4. The power factor is given by:
   \[
   PF = \cos \theta = \frac{P}{\text{VA}}
   \]

These calculations will provide insight into the behavior of the circuit components under AC conditions.
Transcribed Image Text:**Problem Statement:** 1. Assume that the circuit shown in Figure 21-1 is connected to a 480-V, 60-Hz line. The capacitor has a capacitance of 165.782 µF, and the resistor has a resistance of 12 Ω. Find the missing values. **Given Data:** - Voltage, \( E = 480 \) V - Resistance, \( R = 12 \) Ω - Capacitance, \( C = 165.782 \) µF **To Find:** - \( Z \) (Impedance) - \( I_T \) (Total current) - \( E_R \) - \( I_R \) - \( E_C \) - \( I_C \) - \( X_C \) (Capacitive reactance) - \( VARS \) - \( VA \) - \( PF \) (Power factor) - \( P \) (Real power) - \( \theta \) (Phase angle) **Explanation:** The setup involves a resistor and capacitor connected to an AC source. The information provided suggests a need to calculate the collective properties of impedance in the circuit, analyze individual components, and determine both the real and reactive power. **Calculation Guidance:** 1. Use the formulas for capacitive reactance and impedance: \[ X_C = \frac{1}{2 \pi f C} \] \[ Z = \sqrt{R^2 + X_C^2} \] 2. Find the currents and voltages across the resistor and capacitor using Ohm's Law: \[ I_R = \frac{E_R}{R} \] \[ I_C = \frac{E_C}{X_C} \] 3. Calculate power values: \[ P = VI \cos \theta \] \[ VARS = VI \sin \theta \] 4. The power factor is given by: \[ PF = \cos \theta = \frac{P}{\text{VA}} \] These calculations will provide insight into the behavior of the circuit components under AC conditions.
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