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 $30 \mathrm{\Omega }$ 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. $2.0 \mathrm{A}$

In the given circuit the reading of voltmeter $V_1 and V_2$ 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

A condenser of capacity C is charged to a potential difference of ${\mathrm{V}}_1.$ The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to ${\mathrm{V}}_2$ is

(1) ${\left(\frac{\mathrm{C}({\mathrm{V}}_1-{\mathrm{V}}_2{)}^2}{\mathrm{L}}\right)}^{\frac{1}{2}}$                                     

(2) $\frac{\mathrm{C}({\mathrm{V}}_1^2-{\mathrm{V}}_2^2)}{\mathrm{L}}$

(3) $\frac{\mathrm{C}({\mathrm{V}}_1^2+{\mathrm{V}}_2^2)}{\mathrm{L}}$                                             

(4) ${\left(\frac{\mathrm{C}({\mathrm{V}}_1^2-{\mathrm{V}}_2^2)}{\mathrm{L}}\right)}^{\frac{1}{2}}$

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) $E_o{I}_o$

The figure shows a series LCR circuit with L = 5.0 H, C = 80 µF, R = 40 Ω connected to a variable frequency 240 V source. Calculate

(i) The angular frequency of the source which drives the circuit at resonance.

(ii) The current at the resonating frequency.

(iii)The rms potential drop across the capacitor at resonance.                                    3mark