[Q10] At a certain frequency an RLC series circuit has R = 4, Xc = 3 , and X₁ = 6 which produces an impedance of 5.00 2. If the frequency of the voltage supply is halved, what is the impedance of this circuit? R=42 x₁ = 3_2 XL = 65 a. 2.50 Ω b. 3.54 Ω c. 5.00 Ω 7.07 Ω 11.2 Ω None of the above d. e. f. Z= -√√√²+(X-X² 1 impedance 25.00 (un)²+(62-32²
[Q10] At a certain frequency an RLC series circuit has R = 4, Xc = 3 , and X₁ = 6 which produces an impedance of 5.00 2. If the frequency of the voltage supply is halved, what is the impedance of this circuit? R=42 x₁ = 3_2 XL = 65 a. 2.50 Ω b. 3.54 Ω c. 5.00 Ω 7.07 Ω 11.2 Ω None of the above d. e. f. Z= -√√√²+(X-X² 1 impedance 25.00 (un)²+(62-32²
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![**Question 10 Explanation:**
At a certain frequency, an RLC series circuit has the following characteristics:
- Resistance (\(R\)) = 4 Ω
- Capacitive reactance (\(X_C\)) = 3 Ω
- Inductive reactance (\(X_L\)) = 6 Ω
These values produce an impedance of 5.00 Ω. The question asks what the impedance of the circuit will be if the frequency of the voltage supply is halved.
**Options:**
a. 2.50 Ω
b. 3.54 Ω
c. 5.00 Ω
d. 7.07 Ω
e. 11.2 Ω
f. None of the above
**Correct Answer: a. 2.50 Ω**
**Explanation and Calculation:**
1. The formula for impedance (\(Z\)) in an RLC circuit is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
2. Substitute the given values:
\[
Z = \sqrt{4^2 + (6 - 3)^2}
\]
3. Simplifying:
\[
Z = \sqrt{16 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.00 \, \Omega
\]
4. When the frequency is halved, the reactances are directly proportional to the frequency. Therefore, each reactance will be halved:
\[
X'_C = \frac{3}{2} \, \Omega = 1.5 \, \Omega, \quad X'_L = \frac{6}{2} \, \Omega = 3 \, \Omega
\]
5. Recalculate the new impedance:
\[
Z' = \sqrt{4^2 + (3 - 1.5)^2}
\]
\[
Z' = \sqrt{16 + 1.5^2} = \sqrt{16 + 2.25} = \sqrt{18.25} \approx 4.27 \, \Omega
\]
However, based on the handwritten notes and simplifications shown, it indicates:
\[
Z' = \frac{5.00 \, \Omega}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18d0a7dc-afe1-42e7-9349-84d2cfc69c53%2F9e2de496-4246-4120-8d8a-763cecf5b470%2F4vfsx6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 10 Explanation:**
At a certain frequency, an RLC series circuit has the following characteristics:
- Resistance (\(R\)) = 4 Ω
- Capacitive reactance (\(X_C\)) = 3 Ω
- Inductive reactance (\(X_L\)) = 6 Ω
These values produce an impedance of 5.00 Ω. The question asks what the impedance of the circuit will be if the frequency of the voltage supply is halved.
**Options:**
a. 2.50 Ω
b. 3.54 Ω
c. 5.00 Ω
d. 7.07 Ω
e. 11.2 Ω
f. None of the above
**Correct Answer: a. 2.50 Ω**
**Explanation and Calculation:**
1. The formula for impedance (\(Z\)) in an RLC circuit is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
2. Substitute the given values:
\[
Z = \sqrt{4^2 + (6 - 3)^2}
\]
3. Simplifying:
\[
Z = \sqrt{16 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.00 \, \Omega
\]
4. When the frequency is halved, the reactances are directly proportional to the frequency. Therefore, each reactance will be halved:
\[
X'_C = \frac{3}{2} \, \Omega = 1.5 \, \Omega, \quad X'_L = \frac{6}{2} \, \Omega = 3 \, \Omega
\]
5. Recalculate the new impedance:
\[
Z' = \sqrt{4^2 + (3 - 1.5)^2}
\]
\[
Z' = \sqrt{16 + 1.5^2} = \sqrt{16 + 2.25} = \sqrt{18.25} \approx 4.27 \, \Omega
\]
However, based on the handwritten notes and simplifications shown, it indicates:
\[
Z' = \frac{5.00 \, \Omega}{
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