⇒AC Circuits

Correct Option is (B)

$\begin{array}{rl}\tan \varphi & =\tan {45}^{\circ }=\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\\ \therefore \frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}& =1\text{ or }{\mathrm{X}}_{\mathrm{L}}=\mathrm{R}\\ \therefore 2\pi \mathrm{fL}& =\mathrm{R}\\ & \text{ or }\mathrm{L}=\frac{\mathrm{R}}{2\pi \mathrm{f}}=\frac{100}{2\pi \times 1000}=\frac{0.05}{\pi }\mathrm{H}\end{array}$

Correct Option is (D)

$R=500\mathrm{\Omega },\varphi ={60}^{\circ }$

When capacitance is removed

$\begin{array}{rl}& \tan \varphi =\tan {60}^{\circ }=\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\\ \therefore & \frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}=\sqrt{3}\text{ OR }{\mathrm{X}}_{\mathrm{L}}=\mathrm{R}\sqrt{3}\end{array}$

When inductance is removed,

$\begin{array}{rl}& \frac{X_{\mathrm{C}}}{R}=\tan {60}^{\circ }=\sqrt{3}\\ \therefore X_C& =R\sqrt{3}\\ \therefore X_C& =X_L\end{array}$

Impedance $\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}})}^2}=\mathrm{R}=500\mathrm{\Omega }$

Correct Option is (D)

When a d.c. voltage is applied only the resistance of the coil comes into play, its inductive reactance is zero.

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{200}{1}=200\mathrm{\Omega }$

When a.c. voltage is applied, the resistance and inductive reactance come into play and the coil has an impedance Z.

The impedance

$\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{200}{0.5}=400\mathrm{\Omega }$

${\mathrm{Z}}^2={\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2$

$\therefore (400{)}^2=(200{)}^2+{\mathrm{X}}_{\mathrm{L}}^2$

$\therefore {\mathrm{X}}_{\mathrm{L}}^2=(400{)}^2-(200{)}^2=12\times {10}^4$

$\therefore {\mathrm{X}}_{\mathrm{L}}=2\sqrt{3}\times {10}^2=200\sqrt{3}\mathrm{\Omega }$ ..... (1)

Also ${\mathrm{X}}_{\mathrm{L}}=2\pi \mathrm{fL}=2\pi \mathrm{f}\times \frac{2\sqrt{3}}{\pi }=4\sqrt{3}\mathrm{f}$ ..... (2)

By (1) and (2) : $4\sqrt{3}\mathrm{f}=200\sqrt{3}$

$\therefore \mathrm{f}=50\mathrm{Hz}$

Correct Option is (C)

Since the coils are connected to a dc voltage. Inductive reactance will be zero, only resistance has to be considered.

When the coil is broken into two identical points, each part will have resistance $\frac{\mathrm{R}}{2}$. When these are connected in parallel, their equivalent resistance will be $\frac{\mathrm{R}}{4}$. Hence the current I is given by $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}/4}=\frac{4\mathrm{E}}{\mathrm{R}}$

Correct Option is (D)

Inductive reactance when 100 V, 50 Hz source is connected is given by

${\mathrm{X}}_{\mathrm{L}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{8}=12.5\mathrm{\Omega }$

Resistance $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{10}=10\mathrm{\Omega }$

When 100V, 40 Hz supply is connected the inductive reactance will get reduced

Since it is proportional to the frequency

$\begin{array}{rl}& \therefore \frac{{\mathrm{X}}_{\mathrm{L}}^{\prime }}{}=\frac{40}{50}=\frac{4}{5}\\ & \therefore {\mathrm{X}}_{\mathrm{L}}^{\prime }=\frac{4}{5}{\mathrm{X}}_{\mathrm{L}}=\frac{4}{5}\times 12.5=10\mathrm{\Omega }\end{array}$

The resistance will remain same. When they are connected in series the impedance will be given by

$\begin{array}{rl}& Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(10{)}^2+(10{)}^2}=10\sqrt{2}A\\ & I=\frac{V}{Z}=\frac{100}{10\sqrt{2}}=5\sqrt{2}\mathrm{A}\end{array}$