Find the peak current and resonant frequency of the following circuit (as shown in the figure).

  • A
    $0.2 \, A$ and $50 \, Hz$
  • B
    $0.2 \, A$ and $100 \, Hz$
  • C
    $2 \, A$ and $100 \, Hz$
  • D
    $2 \, A$ and $50 \, Hz$

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Given below are two statements $:$ one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$
Assertion $(A) :$ Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube,the tube will be damaged.
Reason $(R):$ By using the choke coil,the voltage across the tube is reduced by a factor $\left(R / \sqrt{R^2+\omega^2 L^2}\right)$,where $\omega$ is the angular frequency of the supply,$R$ is the resistance,and $L$ is the inductance. If the choke coil were not used,the voltage across the tube would be the same as the applied voltage. In the light of the above statements,choose the most appropriate answer from the options given below $:$

Three alternating voltage sources $V_1 = 3 \sin \omega t \text{ V}$,$V_2 = 5 \sin(\omega t + \phi_1) \text{ V}$,and $V_3 = 5 \sin(\omega t - \phi_2) \text{ V}$ are connected in series with a resistor $R = \sqrt{\frac{7}{3}} \, \Omega$ as shown in the figure (where $\phi_1 = 30^\circ$ and $\phi_2 = 127^\circ$). Find the peak current (in Ampere) through the resistor.

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$A$ circuit with an electrical load having impedance $Z$ is connected with an $AC$ source as shown in the diagram. The source voltage varies in time as $V(t) = 300 \sin (400 t) \text{ V}$,where $t$ is time in seconds. List-$I$ shows various options for the load. The possible currents $i(t)$ in the circuit as a function of time are given in List-$II$. Choose the option that describes the correct match between the entries in List-$I$ to those in List-$II$.
List-$I$ List-$II$
$(P)$ Resistor $R = 30 \ \Omega$ $(1)$ $i(t) = 5 \sin(400t)$
$(Q)$ Resistor $R = 30 \ \Omega$ and Inductor $L = 100 \text{ mH}$ $(2)$ $i(t) = 6 \sin(400t + 53^{\circ})$
$(R)$ Capacitor $C = 50 \ \mu\text{F}$,Resistor $R = 30 \ \Omega$,and Inductor $L = 25 \text{ mH}$ $(3)$ $i(t) = 10 \sin(400t)$
$(S)$ Capacitor $C = 50 \ \mu\text{F}$,Resistor $R = 60 \ \Omega$,and Inductor $L = 125 \text{ mH}$ $(4)$ $i(t) = 20 \sin(400t - 90^{\circ})$
$(5)$ $i(t) = 6 \sin(400t - 53^{\circ})$

Is the current distribution shown in the figure possible?

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