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An inductance L having a resistance R is connected to an alternating source of angular frequency ω. The quality factor (Q) of the inductance is L/R
- a)(ωL/R)2
- b)(R/ωL)1∕2
- c)ωL/R
- d)R/ωL
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An inductance L having a resistance R is connected to an alternating s...
Ω. The current in the circuit can be represented by the equation:
I = I0e^(jωt)
where I is the current at any given time, I0 is the amplitude of the current, ω is the angular frequency of the source, and t is time.
In an RL circuit, the voltage across the inductor can be represented by the equation:
V = jωLI
where V is the voltage across the inductor, j is the imaginary unit (√-1), ω is the angular frequency, L is the inductance, and I is the current.
The voltage across the resistor can be represented by the equation:
V = IR
where V is the voltage across the resistor, I is the current, and R is the resistance.
Since the voltage across the inductor and the voltage across the resistor are equal in an RL circuit, we can equate the two equations:
jωLI = IR
We can rearrange the equation to solve for the current:
I = (jωL) / R
The magnitude of the current can be found by taking the absolute value:
|I| = |(jωL) / R|
The phase angle of the current can be found by taking the argument of the complex number:
arg(I) = arg((jωL) / R) = arg(jωL) - arg(R) = π/2 - 0 = π/2
Therefore, the current in the RL circuit lags the voltage by a phase angle of π/2.
I = I0e^(jωt)
where I is the current at any given time, I0 is the amplitude of the current, ω is the angular frequency of the source, and t is time.
In an RL circuit, the voltage across the inductor can be represented by the equation:
V = jωLI
where V is the voltage across the inductor, j is the imaginary unit (√-1), ω is the angular frequency, L is the inductance, and I is the current.
The voltage across the resistor can be represented by the equation:
V = IR
where V is the voltage across the resistor, I is the current, and R is the resistance.
Since the voltage across the inductor and the voltage across the resistor are equal in an RL circuit, we can equate the two equations:
jωLI = IR
We can rearrange the equation to solve for the current:
I = (jωL) / R
The magnitude of the current can be found by taking the absolute value:
|I| = |(jωL) / R|
The phase angle of the current can be found by taking the argument of the complex number:
arg(I) = arg((jωL) / R) = arg(jωL) - arg(R) = π/2 - 0 = π/2
Therefore, the current in the RL circuit lags the voltage by a phase angle of π/2.
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An inductance L having a resistance R is connected to an alternating s...
Option C
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An inductance L having a resistance R is connected to an alternating source of angular frequency ω. The quality factor (Q) of the inductance is L/Ra)(ωL/R)2b)(R/ωL)12c)ωL/Rd)R/ωLCorrect answer is option 'C'. Can you explain this answer?
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