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 (A) \(\cfrac{\pi}{2\sqrt2}\)

An acceptor circuit is a series LCR resonant circuit used in communications and broadcasting to selectively pass a current for a signal of only the desired frequency.

The resonance curve of a series LCR resonant circuit with a small resistance exhibits a very sharp peak at a certain frequency called the resonant frequency f_r. For an alternating signal of this frequency, the impedance of the circuit is minimum, equal to R, and the current is maximum. That is, the circuit has a selective property as it prefers to pass a signal of frequency f_r and reject those of other frequencies.

Use : An acceptor circuit is used in a radio or television receiver to accept the signal of a desired broadcasting station or channel from all the signals that arrive concurrently at its antenna. Tuning a receiver means adjusting the acceptor circuit to be resonant at a desired frequency.

A rejector circuit is a parallel LC resonant circuit used in communications and broadcasting as well as filter circuits to selectively reject a signal of a certain frequency.

The resonance curve of a parallel resonant circuit with a finite resistance of its inductor windings exhibits a sharp minimum at a certain frequency called the resonant frequency f_r. For an alternating signal of this frequency, the impedance of the circuit is maximum and the current is minimum. That is, the circuit has a selective property to reject a signal of frequency f_r while passing those of other frequencies.

Use : A rejector circuit is used at the output stage of a radiowave transmitter.

Characteristics of a parallel LC AC resonance circuit:

1. Resonance occurs when inductive reactance (X = 2πfL) equals capacitive reactance (X_C = 1/2πfC)

Resonant frequency, f_r = 1/2π√(LC)

2. Impedance is maximum. 

3. Current is minimum. 

4. The circuit rejects f_r but allows the current to flow for other frequencies. Hence, it is called a rejector circuit.

(C) the wattless current

Correct option is:  (B) \(\cfrac{1}{2\pi f(2 \pi f L-R)}\)

(C) lags behind the current by phase angle (π/2) rad

For an LR circuit, the impedance,

Z_LR = \(\sqrt{R^2+X_L^2}\), where X_L is the reactance of the inductor.

 Z_LCR = \(\sqrt{R^2+(X_L -X_C)^2}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\cfrac{\pi}{2}\)rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\cfrac{\pi}{2}\)rad. The decrease in net reactance decreases the total impedance (Z_LCR < Z_LR).

The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.

In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.

[Note : In this case, the power factor is zero.]

The reactance of a capacitor is X_C = \(\cfrac{1}{2\pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_C is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

Data : R = 10 Ω, i₀ = 10 mA, i = \(\cfrac T8\)  

V = Ri₀ sin ωt = (10)(10) sin \(\left(\cfrac{2\pi}T.\cfrac{T}8\right)\)

= 100 sin \(\left(\cfrac{\pi}4\right)\) = \(\cfrac{100}{\sqrt2}\) = 70.72 mV

This is the required voltage.

Data: R = 100 Ω, ⁱ₀ = 2A, f = 50 Hz

H = \(\cfrac{ri^2_0}{2f}\) = \(\cfrac{100(2)^2}{2(50)}\) = 4 J

This is the required quantity.

A phasor is a rotating vector that represents a quantity varying sinusoidally with time.

e_av = 0.6365 e₀ = 0.6365(10) = 6.365 V 

Note: In general, when e = e₀ sin ωt, the correspond ing current is j = i sin (ωt + α), where α is the phase difference between emf e and current j. ¿z may be positive or negative or zero.

i₀ is the peak value of the current and i_av (over half cycle)

= \(\cfrac2{\pi}\) i₀ = 0.6365 i₀].

i_av (over half cycle) = \(\cfrac{2\sqrt2}{\pi}\)i_rms

= \(\cfrac{(2)(1.414)}{3.142}\) (3.142) = 2.828 A

[Note: i_av (over half cycle) < i_rms ]

Correct option is (B) 1

(C) R (i_rms)². 2π/ω

Correct option is (C) zero

Correct option is (B) e₀/ωL

Correct option is (A) 2880 W

i_av (over half cycle) = \(\cfrac2{\pi}\) i₀,and i_{rms = \(\cfrac{i_0}{\sqrt2}\)} 

∴ i_av (over half cycle) = \(\left(\cfrac2{\pi}\right)\) \((\sqrt{2i}_{rms})\) = \(\cfrac{2\sqrt2}{\pi}\)i_rms

Correct option is (A) 6.365 A

Correct option is (D) 7.07V

Correct option is (B) in phase

Correct option is (D) zero rad.

Power factor, cos Φ = \(\cfrac R{\sqrt{R^2+(X_L-X_C)^2}}\)

(i) For R >> (X_L – X_C), cos Φ ≅ 1 

(ii) For R >> (X_L – X_e), cos Φ ≅ zero.

Correct option is: (B) 0.707

Correct option is (A) 30 A

Correct option is (D) R/Z.

Correct option is: (C) 5625 W

e = 220 sin \(\cfrac{2 \pi}{T}\left(\cfrac{T}{12}\right)\) = 220 sin \(\left(\cfrac{\pi}6\right)\)

= 220 \(\left(\cfrac{1}6\right)\) = 110 V.

Capacitive reactance = \(\cfrac{1}{2\pi fc}\)

If the frequency (f) of the applied emf is very low, the capacitive reactance (for reasonable value of capacitance C) will be very high and hence the current through the circuit will be very low (for reasonable value of peak emf).

Correct option is: (B) 5\(\sqrt{3/2}\) A 

Correct option is (A) 0.25 H

(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit. 

The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.

When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.

Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.

In a circuit with pure reactance, there is no production of heat.

In LCR series circuit, at current resonance, 

1. reactance is zero and 

2. impedance equals resistance R.

(C) \(\sqrt{R^2+(X_L-X_C)^2}\)

(D) \(\cfrac{2}{\pi}e_0\)