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A series RLC circuit consisting of R = 10 Ω, XL = 20 Ω and XC = 20 Ω, is connected across an AC supply of 100 V (rms). The magnitude and pulse angle (with respect to supply voltage) of the voltage across the induction coil are respectively.
- a)400∠ - 90°V
- b)200∠ + 90°V
- c)200 ; ∠90°V
- d)400∠ + 90°V
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A series RLC circuit consisting of R = 10 Ω, XL= 20 Ω and ...
R = 10Ω, XL = 20 Ω, XC = 20 Ω
V = 100 V (RMS)
XL = XC = 20 Ω
This is the condition of resonance. Hence the circuit is purely Resistive.
Current (I) and applied voltage (V) are in phase and hence power factor will be unity.
I = V/R = 100∠0/10 = 10∠0 A
The voltage across the inductor, VL = I XL
= (10∠0) × (20∠90°)
= 200 ∠90° V
Magnitude of voltage, |VL| = 200 V
The phase of VL = 90°
Phasor diagram:
Most Upvoted Answer
A series RLC circuit consisting of R = 10 Ω, XL= 20 Ω and ...
Given information:
- R = 10 Ω (resistance)
- XL = 20 Ω (inductive reactance)
- XC = 200 Ω (capacitive reactance)
- Vrms = 200 V (rms voltage of the AC supply)
Key formulas:
- Impedance of the series RLC circuit: Z = √(R^2 + (XL - XC)^2)
- Voltage across the capacitor: Vc = Vrms * (XC / Z)
Solution:
Step 1: Calculate the impedance (Z) of the series RLC circuit.
Using the formula: Z = √(R^2 + (XL - XC)^2)
Substituting the given values: Z = √(10^2 + (20 - 200)^2)
Calculating: Z = √(100 + (-180)^2) = √(100 + 32400) = √32500 ≈ 180.277 Ω
Step 2: Calculate the voltage across the capacitor (Vc) using the formula:
Vc = Vrms * (XC / Z)
Substituting the given values: Vc = 200 * (200 / 180.277)
Calculating: Vc ≈ 200 * 1.109 ≈ 221.818 V
Since the question asks for the rms voltage across the capacitor, we need to convert the calculated voltage to rms.
The rms voltage across the capacitor is equal to the calculated voltage divided by the square root of 2 (1.414).
Therefore, Vrms_c = Vc / √2 ≈ 221.818 / 1.414 ≈ 156.977 V
Step 3: Answer the question:
The rms voltage across the capacitor is approximately 156.977 V. None of the given options match with this value, so none of the provided options are correct.
- R = 10 Ω (resistance)
- XL = 20 Ω (inductive reactance)
- XC = 200 Ω (capacitive reactance)
- Vrms = 200 V (rms voltage of the AC supply)
Key formulas:
- Impedance of the series RLC circuit: Z = √(R^2 + (XL - XC)^2)
- Voltage across the capacitor: Vc = Vrms * (XC / Z)
Solution:
Step 1: Calculate the impedance (Z) of the series RLC circuit.
Using the formula: Z = √(R^2 + (XL - XC)^2)
Substituting the given values: Z = √(10^2 + (20 - 200)^2)
Calculating: Z = √(100 + (-180)^2) = √(100 + 32400) = √32500 ≈ 180.277 Ω
Step 2: Calculate the voltage across the capacitor (Vc) using the formula:
Vc = Vrms * (XC / Z)
Substituting the given values: Vc = 200 * (200 / 180.277)
Calculating: Vc ≈ 200 * 1.109 ≈ 221.818 V
Since the question asks for the rms voltage across the capacitor, we need to convert the calculated voltage to rms.
The rms voltage across the capacitor is equal to the calculated voltage divided by the square root of 2 (1.414).
Therefore, Vrms_c = Vc / √2 ≈ 221.818 / 1.414 ≈ 156.977 V
Step 3: Answer the question:
The rms voltage across the capacitor is approximately 156.977 V. None of the given options match with this value, so none of the provided options are correct.
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A series RLC circuit consisting of R = 10 Ω, XL= 20 Ω and XC= 20 Ω, is connected across an AC supply of 100 V (rms). The magnitude and pulse angle (with respect to supply voltage) of the voltage across the induction coil are respectively.a)400∠ - 90°V b)200∠ + 90°Vc)200 ; ∠90°V d)400∠ + 90°VCorrect answer is option 'C'. Can you explain this answer?
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