An alternating voltage $V\sqrt{2}\sin \omega t$ at different frequencies is applied across a capacitor C as shown in the figure. Which of the following graphs between RMS current $(I)$$$ observed by an ammeter and angular frequency $\omega$ of voltage is correct?
          
1.              2. 
3.              4. 

In a series L-C-R circuit, different physical quantities vary with frequency $\omega$. Which of the following curves represent the correct frequency variation of the corresponding quantity?

1. Curve (l) for R

2. Curve (ll) for $X_C$

3. Curve (lll) for $X_L$

4. All of these

For the circuit shown in figure below, the ammeter $A_2$ reads 1.6 A and ammeter $A_3$ read 0.4 A If ${\omega }_0$ is angular frequency and $f$ is frequency of ac, then

$1. {\omega }_0=\frac{4}{\sqrt{LC}}$
$2. f=\frac{2\pi }{\sqrt{LC}}$
$3. The ammeter A_1 reads 1.2 A$
$4. The ammeter A_1 reads 2 A$

If the reading of $V_1$ and $V_3$ are 100 V each, the reading of $V_2$ is

1. 0 V

2. 100 V

3. 200 V

4. Cannot be determined by given information

Given that the current \(i_1=3A \sin \omega t\) and the current \(i_2=4A \cos \omega t,\) what will be the expression for the current \(i_3\)?

                  
1. \(5 A \sin \left(\omega t+53^{\circ}\right) \)
2. \(5 A \sin \left(\omega t+37^{\circ}\right) \)
3. \(5 A \sin \left(\omega t+45^{\circ}\right) \)
4. \( 5 A \sin \left(\omega t+30^{\circ}\right)\)

A capacitor of capacitance \(1~\mu\text{F}\) is charged to a potential of \(1\) V. It is connected in parallel to an inductor of inductance \(10^{-3}~\text{H}\). What is the value of the maximum current that will flow in the circuit?
1. \(\sqrt{1000}~\text{mA}\)
2. \(1~\text{mA}\)
3. \(1~\mu\text{F}\)
4. \(1000~\text{mA}\)

In a box \(Z\) of unknown elements (\(L\) or \(R\) or any other combination), an ac voltage \(E = E_0 \sin(\omega t + \phi)\) $$is applied and the current in the circuit is found to be \(I = I_0 \sin\left(\omega t + \phi +\frac{\pi}{4}\right)\). The unknown elements in the box could be:
             

In the transformer shown in the figure, the ratio of the number of turns of primary to the secondary is $\frac{N_1}{N_2}=\frac{1}{50}$. If a battery of emf 10 V is connected across primary, then induced current through the load of 10 k$\Omega$ in the secondary is

$1. \frac{1}{20}A$
$2. Zero$
$3. \frac{1}{10}A$
$4. \frac{1}{5}A$