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A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V?
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A coil having a resistance of 5 and inductance of 0.1 H is connected i...
Given:
Resistance (R) = 5 Ω
Inductance (L) = 0.1 H
Capacitance (C) = 50 µF = 50 × 10⁻⁶ F
Applied voltage (V) = 200 V
To find: The voltage across the coil at resonance
Formula used:
Resonance frequency (f) = 1 / (2π√LC)
Impedance (Z) = √(R² + (XL - XC)²)
Voltage (V) = I × Z
Solution:
1. Resonance frequency:
f = 1 / (2π√LC)
= 1 / (2π√0.1 × 50 × 10⁻⁶)
= 1000 Hz
2. Impedance at resonance:
At resonance, XL = XC
Z = √(R² + (XL - XC)²)
= √(5² + (2πfL - 1/(2πfC))²) [Substituting values]
= √(5² + (2π × 1000 × 0.1 - 1/(2π × 1000 × 50 × 10⁻⁶))²)
= √(25 + 398.47²)
= 398.5 Ω (approx.)
3. Current at resonance:
At resonance, the current (I) will be maximum.
I = V / Z
= 200 / 398.5
= 0.502 A (approx.)
4. Voltage across the coil at resonance:
The voltage across the coil (VL) will be equal to the product of current and inductive reactance (XL).
VL = I × XL
= I × 2πfL
= 0.502 × 2π × 1000 × 0.1
= 31.42 V (approx.)
Therefore, the voltage across the coil at resonance is 31.42 V (approx.), which is option (A) 200 V.
Resistance (R) = 5 Ω
Inductance (L) = 0.1 H
Capacitance (C) = 50 µF = 50 × 10⁻⁶ F
Applied voltage (V) = 200 V
To find: The voltage across the coil at resonance
Formula used:
Resonance frequency (f) = 1 / (2π√LC)
Impedance (Z) = √(R² + (XL - XC)²)
Voltage (V) = I × Z
Solution:
1. Resonance frequency:
f = 1 / (2π√LC)
= 1 / (2π√0.1 × 50 × 10⁻⁶)
= 1000 Hz
2. Impedance at resonance:
At resonance, XL = XC
Z = √(R² + (XL - XC)²)
= √(5² + (2πfL - 1/(2πfC))²) [Substituting values]
= √(5² + (2π × 1000 × 0.1 - 1/(2π × 1000 × 50 × 10⁻⁶))²)
= √(25 + 398.47²)
= 398.5 Ω (approx.)
3. Current at resonance:
At resonance, the current (I) will be maximum.
I = V / Z
= 200 / 398.5
= 0.502 A (approx.)
4. Voltage across the coil at resonance:
The voltage across the coil (VL) will be equal to the product of current and inductive reactance (XL).
VL = I × XL
= I × 2πfL
= 0.502 × 2π × 1000 × 0.1
= 31.42 V (approx.)
Therefore, the voltage across the coil at resonance is 31.42 V (approx.), which is option (A) 200 V.
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A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V?
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A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V?.
A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil having a resistance of 5 and inductance of 0.1 H is connected in series with a condenser of capacitance 50 µF. A constant alternating voltage of 200 V is applied to the circuit. The voltage across the coil at resonance is : (A) 200 V (B)1788 V (C) 1800 V (D) 2000 V?.
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