ALTERNATING CURRENT

UG NEET-ALTERNATING CURRENT

What is the r.m.s.value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of 2 amperes in the same resistor :-

6 amp 3.46 amp 2 amp 0.66 amp

The peak value of an alternating e.m.f. which is given by E = E_o cos ωt is 10 volts and its frequency is 50 Hz. At time t = 1/600 s, the instantaneous e.m.f. is

10 V 5 √3 V 5 V 1 V

The phase difference between current and voltage in an AC circuit is π/4 radian, If the frequency of AC is 50 Hz, then the phase difference is equivalent to the time difference:-

0.78 s 15.7 ms 2.5 s 2.5 ms

A current in circuit is given by i = 3 + 4 sin wt. Then the effective value of current is:

5 √7 √10 √17

Incorrect statement are:
(a) A.C. meters can measure D.C also (b) If A.C. meter measures D.C. there scale must be linear and uniform (c) A.C. and D.C. meters are based on heating effect of current (d) A.C. meter reads rms value of current

a,b b,c c,d d,a

The r.m.s. value of current for a variable current i = i₁ cos ωt + i₂ sin ωt :-

$\frac{1}{\sqrt{2}}\left ( i_{1}+i_{2} \right )$ $\frac{1}{\sqrt{2}}\left ( i_{1}+i_{2} \right )^{2}$ $\frac{1}{\sqrt{2}}\left ( i_{1}^{2}+i_{2}^{2} \right )^{1/2}$ $\frac{1}{2}\left ( i_{1}^{2}+i_{2}^{2} \right )^{1/2}$

The relation between an A.C. voltage source and time in SI units is: V = 120 sin (100 πt) cos (100 πt) volt value of peak voltage and frequency will be respectively :-

120/√2 volt and 100 Hz 60 volt and 100 Hz 120 volt and 100 Hz 60 volt and 200 Hz

If an A.C. main supply is given to be 220 V. What would be the average e.m.f. during a positive half cycle :-

386 V 198 V 256 V None of above

The hot wire ammeter measures :-

None of above A.C. current both (1) & (2) D.C. current

Frequency of A.C. in India is-

60 Hz 45 Hz 50 Hz None of the above

For an alterating current I = I_ocos ωt, What is the rms value and peak value of current :-

$\frac{I_{0}}{\sqrt{2}},I_{0}$ $2I_{0},\frac{I_{0}}{\sqrt{2}}$ $I_{0},\frac{I_{0}}{2}$ $I_{0},\frac{I_{0}}{\sqrt{2}}$

If a step up transformer have turn ratio 5, frequency 50 Hz root mean square value of potential difference on primary 100 volts and the resistance of the secondary winding is 500 Ω then the peak value of voltage in secondary winding will be (the efficiency of the transformer is hundred percent)

10√2 50√2 500√2 20√2

A resonant A.C. circuit contains a capacitor of capacitance 10 ^-6 F and an inductor of 10 ^-4 H. The frequency of electrical oscillations will be :-

$\frac{10^{5}}{2\pi } Hz$ $\frac{10}{2\pi } Hz$ 10 Hz 10⁵ Hz

A resistance of 300 Ω and an inductance of 1/π henry are connected in series to a A.C. voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is :-

$tan^{-1}\left ( \frac{2}{3} \right )$ $tan^{-1}\left ( \frac{3}{2} \right )$ $tan^{-1}\left ( \frac{4}{3} \right )$ $tan^{-1}\left ( \frac{3}{4} \right )$

The graphs given below depict the dependence of two reactive impedances X ₁, and X₂ on the frequency of the alternating e.m.f. applied individually to them. We can then say that:

X₁ is a capacitor and X₂ is an inductor X₁ is an inductor and X₂ is a resistor X₁ is an inductor and X₂ is a capacitor X₁ is a resistor and X₂ is a capacitor

A 12 ohm resistor and a 0.21 henry inductor are connected in series to an AC source operating at 20 volts, 50 cycle/second. The phase angle between the current and the source voltage is:

30 ⁰ 80 ⁰ 40 ⁰ 90 ⁰

A 110 V, 60 W lamp is run from a 220 V AC mains using a capacitor in series with the lamp, instead of a resistor then the voltage across the capacitor is about:-

110 V 190 V 220 V 311 V

The resistance that must be connected in series with inductance of 0.2 H in order that the phase difference between current and e.m.f. may be 45^o when the frequency is 50 Hz, is:-

6.28 ohm 62.8 ohm 628 ohm 31.4 ohm

An inductive circuit contains resistance of 10 ohms and an inductance of 20 H. If an A.C. voltage of 120 volt and frequency 60 Hz is applied to this circuit, the current would be nearly

0.80 amp 0.48 amp 0.16 amp 0.016 amp

A student connects a long air cored coil of manganin wire to a 100 V D.C. supply and records a current of 25 amp. When the same coil is connected across 100 V. 50 Hz a.c. the current reduces to 20 A, the reactance of the coil is :-

5 Ω 4 Ω 3Ω None

The graph shows variation of I with f for a series R-L-C network. Keeping L and C constant. If R decreases:

(a)Maximum current (I_m) increases (b)Sharpness of the graph increases (c)Quality factor increases (d)Band width increases

c, d, a b, c, d All a, b, c

Alternating current is flowing in inductance L and resistance R. The frequency of source is ω/2π. Which of the following statement is correct:

For high frequency the limiting value of impedance is ωL. For low frequency the limiting value of impedance is ωL. For high frequency the limiting value of impedance is R. For low frequency the limiting value of impedance is L.

A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then

Bulb will give more intense light. Bulb will give light of same intensity as before Bulb will stop radiating light. Bulb will give less intense light.

In an A.C. circuit resistance and inductance are connected in series. The potential and current in inductance is:

$V_{0} sin \left ( \omega t+\pi /2 \right ),\frac{V_{0}}{\omega L}sin \omega t$ $V_{0} sin \omega t,\frac{V_{0}}{\omega L}sin \omega t$ $V_{0} sin \omega t,\frac{V_{0}}{\omega L}sin \left ( \omega t+\pi /2 \right )$ $V_{0} sin \left ( \omega t+\pi /2 \right ),\frac{V_{0}}{\omega L}sin \left ( \omega t-\pi /2 \right )$

An a.c. source of voltage V and of frequency 50 Hz is connected to an inductor of 2 H and negligible resistance. A current of r.m.s value I flows in the coil. When the frequency of the voltage is changed to 400 Hz keeping the magnitude of V the same, the current is now :-

I/4 and lagging by 90⁰ from V 41 and leading by 90⁰ from V I/8 and lagging by 90⁰ from V 8 I in phase with V

A capacitor of capacity C is connected in A.C. circuit. The applied emf is V = V₀ sin ωt, then the current is:

$I = \frac{V_{0}}{\omega L}sin\omega t$ $I = \frac{V_{0}}{\omega L}sin\left ( \omega t+\pi /2 \right )$ $I = V_{0}\omega C sin \omega t$ $I = V_{0}\omega C sin \left ( \omega t+\pi /2 \right )$

The impedence of a circuit, when a resistance R and an inductor of inductance L are connected in series in an A.C. circuit of frequency (f) is :-

$\sqrt{R+4\pi fL^{2}}$ $\sqrt{R+4\pi^{2} f^{2}L^{2}}$ $\sqrt{R^{2}+2\pi^{2} f^{2}L^{2}}$ $\sqrt{R^{2}+4\pi^{2} f^{2}L^{2}}$

A capacitor of capacity C and reactance X if capacitance and frequency become double then reactance will be :-

2X X/2 4X X/4

The coil of choke in a circuit :

increases the current has high resistance to d.c. circuit does not change the current controlled the current

The inductive reactance of an inductive coil with 1/π henry and 50 Hz :-

π/50 ohm 50/π ohm 50 ohm 100 ohm

In the L-R circuit R = 10 Ω and L = 2H. If 120 V, 60 Hz alternating voltage is applied then the flowing current in this circuit will be :-

0.80 A 0.48 A 0.32 A 0.16 A

An Inductance of 0.4 Henry and a resistance of 100 ohm are connected to a A.C. voltage source of 220 V and 50 Hz. Then find out the phase difference between the voltage and current flowing in the circuit :

$tan^{-1}(0.5 \pi )$ $tan^{-1}(1.5 \pi )$ $tan^{-1}(2.25 \pi )$ $tan^{-1}(0.4 \pi )$

A capacitor of capacitance 100 μF & a resistance of 100 Ω is connected in series with AC supply of 220V, 50Hz. The current leads the voltage by

$tan^{-1}\left ( \frac{2}{\pi} \right ) $ $tan^{-1}\left ( \frac{1}{2\pi} \right ) $ $tan^{-1}\left ( \frac{4}{\pi} \right ) $ $tan^{-1}\left ( \frac{1}{\pi} \right ) $

If the current through an inductor of inductance L is given by I = I₀ sin ωt, then the voltage across inductor will be :-

$I_{0}\omega Lsin(\omega t-\pi/2)$ $I_{0}\omega Lsin(\omega t-\pi)$ None of these $I_{0}\omega Lsin(\omega t+\pi/2)$

There is a 5 Ω resistance in an A.C., circuit. Inductance of 0.1 H is connected with it in series. If equation of A.C. e.m.f. is 5 sin 50 t then the phase difference between current and e.m.f. is :-

0 π/6 π/2 π/4

A 50 Hz a.c. source of 20 volts is connected across R and C as shown in figure below. The voltage across R is 12 volts. The voltage across C is

10 V 8 V Not possible to determine unless values of R and C are given 16 V

200 Ω resistance and 1H inductance are connected in series with an A.C. circuit. The frequency of the source is 200/2π Hz. Then phase difference in between V and I will be :-

30⁰ 60⁰ 90⁰ 45⁰

Impedance of the following circuit will be:

150 Ω 200 Ω 340 Ω 250 Ω

In showing figure find V_R:

√220 x 176 V 185 V 396 V 132 V

If alternating current of 60 Hz frequency is flowing through inductance of L = 1 mH and drop in ∆V_L is 0.6 V then alterating current :-

20/π A 50/π A 1/ π A 5/π A

An inductance of 1mH, a condenser of 10μFand a resistance of 50 Ω are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be :-

100 Ω 30 Ω 3.2 Ω 10 Ω

L, Cand R represent physical quantities inductance, capacitance and resistance respectively. The combination representing dimension of frequency is

LC $\left ( \frac{L}{C} \right )^{-1/2}$ $\frac{C}{L}$ $\left ( LC \right )^{-1/2}$

A circuit contains R, L and C connected in series with an A. C. source. The values of the reactances for inductor and capacitor are 200 Ω and 600 Ω respectively and the impedance of the circuit is Z₁. What happens to the impedance of the same circuit if the values of the reactances are interchanged: -

The impedance will decrease Information insufficient The impedance will increase The impedance will remain unchanged

When V = 100 sinωt is applied across a series (R- L-C) circuit, At resonance the current in resistance (R = 100 Ω) is I = i₀, sinωt, then power dissipation in circuit is:-

25 W Can't be calculated 100 W 50 W

At resosnance in a series LCR circuit, which of the following statements is true:-

Current in the circuit is maximum and phase difference between E and I is π/2 Voltage is maximum and phase difference between E and I is π/2 Current is minimum and phase difference between E and I is zero Current in the circuit is maximum and phase difference between E and I is zero

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