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

Power dissipated in an L-C-R series circuit connected to an AC source of emf $\epsilon$ is 

(a) $\frac{{\epsilon }^{2}R}{\left[{\displaystyle R^2+{\left(L\omega -\frac{1}{C\omega }\right)}^2}\right]}$

(b) $\frac{{\epsilon }^2{\sqrt{R^2+\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}}^2}{R}$

(c) $\frac{{\epsilon }^2\left[R^2+{\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}^2\right]}{R}$

(d) $\frac{{\epsilon }^{2}R}{\sqrt{R^2+{\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}^2}}$

In AC circuit the emf (e) and the current (i) at any instant are given respectively by

$e=E_o \sin \omega t$

$i=I_o\sin (\omega t-\varphi )$

The average power in the circuit over one cycle of AC is 

1.$\frac{E_o{I}_o}{2}$

2. $\frac{E_o{I}_o}{2}\sin \varphi$

3. $\frac{E_o{I}_o}{2}\cos \varphi$

4. $E_o{I}_o$

A coil has resistance $30 \mathrm{\Omega }$ and inductive reactance $20 \mathrm{\Omega }$ 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. $4.0 \mathrm{A}$                                          

 2. $8.0 \mathrm{A}$

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

 4. $2.0 \mathrm{A}$

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

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 instantaneous values of alternating

current and voltages in a circuit are given

as 

$\mathrm{i}=\frac{1}{\sqrt{2}}\sin \left(100\mathrm{\pi t}\right) \mathrm{ampere}$
$\mathrm{e}=\frac{1}{\sqrt{2}}\sin (100\mathrm{\pi t}+\mathrm{\pi }/3) \mathrm{volt}$

The average power in Watts consumed in the

circuit is

(a) $\frac{1}{4}$

(b) $\frac{\sqrt{3}}{4}$

(c) $\frac{1}{2}$ 

(d) $\frac{1}{8}$

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) $\sqrt{3}/2$

A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively are
1. 300V,15A
2. 450V,15A

3. 450V,13.5A

4. 600V,15A