Alternating Current
- 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
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Quality Factor = Potential drop accross capacitor or inductor Potential drop accross R = IωL = ωL IR R
Correct Option: D
Quality Factor = Potential drop accross capacitor or inductor Potential drop accross R = IωL = ωL IR R
- A capacitor has capacity C and reactance X. If capacitance
and frequency become double, then reactance will be
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Capacitive reactance,
X = 1 = 1 ωC 2πƒC ⇒ X ∝ 1 ƒ C ∴ X' = ƒ × C ƒ × C = 1 X ƒ' C' 2ƒ 2C 4 ⇒ X' = X 4
Correct Option: C
Capacitive reactance,
X = 1 = 1 ωC 2πƒC ⇒ X ∝ 1 ƒ C ∴ X' = ƒ × C ƒ × C = 1 X ƒ' C' 2ƒ 2C 4 ⇒ X' = X 4
- In a series resonant circuit, having L, C and R as its elements, the resonant current is i. The power dissipated in the circuit at resonance is
where ω is the angular resonance frequency.
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At resonance Lω = 1 ,ω Cω = 1 √LC Current through circuit i = E R
Power dissipated at Resonance = i²RCorrect Option: D
At resonance Lω = 1 ,ω Cω = 1 √LC Current through circuit i = E R
Power dissipated at Resonance = i²R
- A coil of 40 henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
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Time constant is L/R
Given, L = 40H & R = 8Ω
∴ τ = 40/8 = 5 sec.Correct Option: B
Time constant is L/R
Given, L = 40H & R = 8Ω
∴ τ = 40/8 = 5 sec.
- In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is
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From figure,
tan 45º = (1/ωC) - ωL R ⇒ 1 - ωL = R ωC ⇒ 1 = R + ωL ωC ∴ ω = 2π = 2μƒ T C = 1 = 1 ω(R + ωL) 2πƒ(R + 2πƒ L)
Correct Option: D
From figure,
tan 45º = (1/ωC) - ωL R ⇒ 1 - ωL = R ωC ⇒ 1 = R + ωL ωC ∴ ω = 2π = 2μƒ T C = 1 = 1 ω(R + ωL) 2πƒ(R + 2πƒ L)