A coil of self-inductance \(L\) is connected in series with a bulb \(B\) and an AC source. The brightness of the bulb decreases when:

1.	Frequency of the AC source is decreased
2.	The number of turns in the coil is reduced
3.	A capacitance of reactance \(X_C = X_L\) is included in the same circuit
4.	An iron rod is inserted in the coil

In an electrical circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\mathrm{\pi }/3.$ If instead, C is removed from the circuit, the phase difference is again $\mathrm{\pi }/3.$ The power factor of the circuit is

1. 1/2                                           

2. 1/$\sqrt{2}$

3. 1                                             

4. $\sqrt{3}/2$

The instantaneous values of alternating

current and voltages in a circuit are given

as 

$\mathrm{i}=\frac{1}{\sqrt{2}}\sin \left(100\mathrm{\pi t}\right)<mspace\; width="1em"></mspace>\mathrm{ampere}$
$\mathrm{e}=\frac{1}{\sqrt{2}}\sin (100\mathrm{\pi t}+\mathrm{\pi }/3)<mspace\; width="1em"></mspace>\mathrm{volt}$

The average power in Watts consumed in the

circuit is

1. $\frac{1}{4}$

2. $\frac{\sqrt{3}}{4}$

3. $\frac{1}{2}$ 

4. $\frac{1}{8}$

In an AC circuit an alternating voltage $\mathrm{e}=200<mspace\; width="1em"></mspace>\sqrt{2}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>100\mathrm{t}$ volt is connected to a capacitor $1<mspace\; width="1em"></mspace>\mathrm{\mu F}.$ The $\mathrm{rms}$ value of the current in the circuit is

(a) 100 mA

(b) 200 mA

(c) 20 mA

(d) 10 mA

The current i  in a coil varies with time as shown in the figure. The variation of induced emf with time would be

                  (a)                                                                                         (b)

                   (c)                                                                                             (d)

An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3~ \Omega, \) then the phase difference between the applied voltage and the current in the circuit will be:

The rms value of potential difference V shown in the figure is

(1) ${\mathrm{V}}_{\mathrm{o}}$                                                     

(2) $\frac{{\mathrm{V}}_{\mathrm{o}}}{\sqrt{2}}$

(3)$\frac{{\mathrm{V}}_{\mathrm{o}}}{2}$                                                     

(4) $\frac{{\mathrm{V}}_{\mathrm{o}}}{\sqrt{3}}$   

A coil has resistance $30<mspace\; width="1em"></mspace>\mathrm{\Omega }$ and inductive reactance $20<mspace\; width="1em"></mspace>\mathrm{\Omega }$ at 50 Hz frequency. If an AC source of 200 V, 100 Hz, is connected across the coil, the current in the coil will be:

 1. $4.0<mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace>$                                          

 2. $8.0<mspace\; width="1em"></mspace>\mathrm{A}$

 3. $\frac{20}{\sqrt{13}}\mathrm{A}$                                         

 4. $2.0<mspace\; width="1em"></mspace>\mathrm{A}$

In the given circuit the reading of voltmeter $V_1<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>V_2$ are 300V each. The reading to the voltmeter $V_3$ and ammeter A are respectively :

(a) 150V, 2.2A                                             

(b) 220V, 2.2A

(c) 220V, 2.0A                                             

(d) 100V, 2.0A

A 220V input is supplied to a transformer.The output circuit draws a current of 2.0A at 440V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is 

1. 3.6A                                        2. 2.8A

3. 2.5A                                        4. 5.0A