The current I, potential difference V_L across the inductor and potential difference V_C across the capacitor in the circuit as shown in the figure are best represented vectorially as:

                     

1.                                      2.  

3.                                  4.  

Resonance occurs in a series L-C-R circuit when the frequency of the applied e.m.f. is 1000 Hz. Then:

1.  when f=900 Hz, the circuit behaves as a capacitive circuit

2.  the impendence of the circuit is maximum at f=1000 Hz

3.  at resonance the voltages across L and current in C differ in phase by 180$°$

4.  if the value of C is doubled resonance occurs at f=2000 Hz

A series combination of R, L, C is connected to an a.c. source. If the resistance is 3 $\Omega$ and the reactance is 4 $\Omega$, the power factor of the circuit is

(1) 0.4                  (2) 0.6                   (3) 0.8               (4) 1.0

A coil has an inductance of 0.7 H and is joined in series with a resistance of 220 $\Omega$. An alternating emf of 220 V at 50 Hz is applied to it. Then the wattless component of the current in the circuit is

(1) 5 amp         (2) 0.5 amp       (3) 0.7 amp        (4) 7 amp

The voltage and current in a series AC circuit are given by V = $V_0$ cos $\omega t and I = I_0 (\cos \omega t+\frac{\mathrm{\pi }}{3}).$  What is the power dissipated in the circuit?

(1) $\frac{V_0{I}_0}{2}$                (2) 0              (3) $\frac{V_0{I}_0}{4}$         (4) $\frac{\sqrt{3}}{2}{V}_0{I}_0$

In the circuit of figure, what will be the reading of the voltmeter?

(1) 300 V           (2) 900 V

(3) 200 V         (4) 400 V

The root-mean-square value of an alternating current of \(50~\text{Hz}\) frequency is \(10 ~\text A\). The time taken by  the alternating current to reach from zero to maximum value and peak value will be:
1. \(2×10^{–2}~\text s\) and \(14.14 ~\text A \)
2. \(2×10^{–2}~\text s\) and \(7.07~\text A\)
3. \(5×10^{–3}~\text s\) and \(7.07~\text A\)
4. \(5×10^{–3}~\text s\) and \(14.14 ~\text A \)

A 120 volt AC source is connected across a pure inductor of inductance 0.70 Henry. If the frequency of the source is 60 Hz, the current passing through the inductor is :

(1) 4.55 A                       (2) 0.325 A

(3) 0.455 A                     (4) 3.25 A

In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

(1) 4 L                (2) 2 L

(3) L/2              (4) L/4

In a series R-L-C AC circuit, for a particular value of R, L and C, power supplied by the source is P at resonance. If the  value of inductance is halved, then the power from the source again at resonance is P'. Then

(1) P = $\frac{P\text{'}}{2}$

(2) P = 2P'

(3) P = 4P'

(4) P = P'