The inductive reactance and resistance of the L-R circuit are respectively 30 ohm and 40 ohm. The impedance of the circuit is

1. 50 ohm

2. 10 ohm

3. 70 ohm

4. $\frac{4}{3}$ ohm

Voltage across the resistor, inductor and capacitor in series LCR circuit are 20 V, 50 V and 50 V respectively. Peak value of the applied source is

1. 20 V

2. 20$\sqrt{2}$ V

3. 120$\sqrt{2}$ V

4. 120 V

The r.m.s value of current for

$l=<mspace\; width="1em"></mspace>3\sin \omega t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>\cos \omega t<mspace\; width="1em"></mspace>is<mspace\; width="1em"></mspace>$

1. 5 A

2. 5$\sqrt{2}$ A

3. $\frac{5}{\sqrt{2}}A$

4. $\frac{7}{\sqrt{2}}A$

The r.m.s. voltage of the waveform shown is:

1. 10 V

2. 6.37 V

3. 7 V

4. 10.5 v

Voltage and current in an A.C. circuit is given by $V=20\sin \left(314t\right)<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>l=10\sin \left(314t+\frac{\pi }{3}\right)<mspace\; width="1em"></mspace>A.$

Wattful current in the circuit is-

$1.<mspace\; width="1em"></mspace>10<mspace\; width="1em"></mspace>A<mspace\; width="1em"></mspace>$
$2.<mspace\; width="1em"></mspace>5<mspace\; width="1em"></mspace>A<mspace\; width="1em"></mspace>$
$3.<mspace\; width="1em"></mspace>\frac{5}{\sqrt{2}}A<mspace\; width="1em"></mspace>$
$4.<mspace\; width="1em"></mspace>\frac{10}{\sqrt{2}}A<mspace\; width="1em"></mspace>$

A time-varying current is given by $l=5\sin \left(\omega t+\begin{array}{c}\pi \\ 6\end{array}\right)$ Its r.m.s. value is (symbols have usual meanings)

$1.<mspace\; width="1em"></mspace>5\sqrt{5}<mspace\; width="1em"></mspace>A$
$2.<mspace\; width="1em"></mspace>10<mspace\; width="1em"></mspace>A<mspace\; width="1em"></mspace>$
$3.<mspace\; width="1em"></mspace>\frac{5}{\sqrt{2}}<mspace\; width="1em"></mspace>A<mspace\; width="1em"></mspace>$
$4.<mspace\; width="1em"></mspace>\frac{15}{\sqrt{2}}A<mspace\; width="1em"></mspace>$

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

$1.<mspace\; width="1em"></mspace>{\omega }_0=\frac{4}{\sqrt{LC}}$
$2.<mspace\; width="1em"></mspace>f=\frac{2\pi }{\sqrt{LC}}$
$3.<mspace\; width="1em"></mspace>The<mspace\; width="1em"></mspace>ammeter<mspace\; width="1em"></mspace>A_1<mspace\; width="1em"></mspace>reads<mspace\; width="1em"></mspace>1.2<mspace\; width="1em"></mspace>A<mspace\; width="1em"></mspace>$
$4.<mspace\; width="1em"></mspace>The<mspace\; width="1em"></mspace>ammeter<mspace\; width="1em"></mspace>A_1<mspace\; width="1em"></mspace>reads<mspace\; width="1em"></mspace>2<mspace\; width="1em"></mspace>A$

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. \( 5 A \sin \left(\omega t+30^{\circ}\right)\)