10. The resistor of an RLC series circuit has a resistance of 48 Q. The inductor has an induc- tive reactance of 88 2 and the capacitor has a capacitive reactance of 62 2. The apparent power of the circuit is 780 VA. What is the total current in the circuit?
10. The resistor of an RLC series circuit has a resistance of 48 Q. The inductor has an induc- tive reactance of 88 2 and the capacitor has a capacitive reactance of 62 2. The apparent power of the circuit is 780 VA. What is the total current in the circuit?
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![**Problem 10: RLC Circuit Analysis**
The resistor of an RLC series circuit has a resistance of 48 Ω. The inductor has an inductive reactance of 88 Ω, and the capacitor has a capacitive reactance of 62 Ω. The apparent power of the circuit is 780 VA. What is the total current in the circuit?
**Analysis:**
To find the total current in the circuit, we will first need to determine the impedance (Z) of the RLC series circuit. The impedance in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( R = 48 \, \Omega \) (resistance)
- \( X_L = 88 \, \Omega \) (inductive reactance)
- \( X_C = 62 \, \Omega \) (capacitive reactance)
Calculate the net reactance (\( X \)):
\[ X = X_L - X_C = 88 \, \Omega - 62 \, \Omega = 26 \, \Omega \]
Now, substitute into the impedance formula:
\[ Z = \sqrt{48^2 + 26^2} \]
\[ Z = \sqrt{2304 + 676} \]
\[ Z = \sqrt{2980} \]
\[ Z \approx 54.6 \, \Omega \]
Next, use the formula for apparent power (S), which is related to the current (I) and impedance (Z):
\[ S = V_{\text{rms}} \times I_{\text{rms}} \]
\[ S = I^2 \times Z \]
Given that the apparent power \( S = 780 \, \text{VA} \), the equation becomes:
\[ 780 = I^2 \times 54.6 \]
Solve for \( I \):
\[ I^2 = \frac{780}{54.6} \]
\[ I^2 \approx 14.29 \]
\[ I \approx \sqrt{14.29} \]
\[ I \approx 3.78 \, \text{A} \]
Thus, the total current in the circuit is approximately \( 3.78 \, \text{A} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c64c691-4381-4c3b-ba72-71ddf7a7103d%2F7976662c-870b-4916-b223-31518e774856%2Fm9xl80m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 10: RLC Circuit Analysis**
The resistor of an RLC series circuit has a resistance of 48 Ω. The inductor has an inductive reactance of 88 Ω, and the capacitor has a capacitive reactance of 62 Ω. The apparent power of the circuit is 780 VA. What is the total current in the circuit?
**Analysis:**
To find the total current in the circuit, we will first need to determine the impedance (Z) of the RLC series circuit. The impedance in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( R = 48 \, \Omega \) (resistance)
- \( X_L = 88 \, \Omega \) (inductive reactance)
- \( X_C = 62 \, \Omega \) (capacitive reactance)
Calculate the net reactance (\( X \)):
\[ X = X_L - X_C = 88 \, \Omega - 62 \, \Omega = 26 \, \Omega \]
Now, substitute into the impedance formula:
\[ Z = \sqrt{48^2 + 26^2} \]
\[ Z = \sqrt{2304 + 676} \]
\[ Z = \sqrt{2980} \]
\[ Z \approx 54.6 \, \Omega \]
Next, use the formula for apparent power (S), which is related to the current (I) and impedance (Z):
\[ S = V_{\text{rms}} \times I_{\text{rms}} \]
\[ S = I^2 \times Z \]
Given that the apparent power \( S = 780 \, \text{VA} \), the equation becomes:
\[ 780 = I^2 \times 54.6 \]
Solve for \( I \):
\[ I^2 = \frac{780}{54.6} \]
\[ I^2 \approx 14.29 \]
\[ I \approx \sqrt{14.29} \]
\[ I \approx 3.78 \, \text{A} \]
Thus, the total current in the circuit is approximately \( 3.78 \, \text{A} \).
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