Alternating Current

$\text { Given, turn ratio of a transformer, } \frac{N_1}{N_2}=\frac{50}{1}$

$\Rightarrow \quad N_1=50 N_2$

$\text { Since, } \quad \frac{N_1}{N_2}=\frac{V_1}{V_2}$

$\begin{array}{rlrl} & 50=\frac{120}{V_2} \quad\left[V_1=120 \mathrm{~V} \text { (given) }\right] \\ \Rightarrow & V_2=\frac{12}{5} \mathrm{~V} \end{array}$

Output power at secondary coil,

$\begin{aligned} & P_s=\frac{V_2^2}{R_2} \quad \quad \quad\left[R_2=1 \Omega \text { (given) }\right] \\ & =\frac{\left(\frac{12}{5}\right)^2}{1}=\frac{144}{25}=5.76 \mathrm{~W} \end{aligned}$

Given the values :

$ L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} $

$ C = 10 \, \mu\text{F} = 10^{-5} \, \text{F} $

The time period $ T $ for an LC oscillation can be calculated using the formula :

$ T = 2\pi \sqrt{LC} $

Substituting the given values :

$ T = 2\pi \sqrt{25 \times 10^{-3} \times 10^{-5}} = \pi \times 10^{-3} \, \text{s} = \pi \, \text{ms} $

For the current in the circuit to be at its maximum, it occurs at :

$ t = \frac{T}{4} = \frac{\pi}{4} \, \text{ms} $

Therefore, the time after which the current in the circuit will be maximum is $\frac{\pi}{4} \, \text{ms}$.

Given, in $L C R$ series circuit,

Source voltage (resultant voltage) $V=120 \mathrm{~V}$

Voltage across inductor, $V_L=50 \mathrm{~V}$

Voltage across resistor, $V_R=40 \mathrm{~V}$

In $L$-$C$-$R$ series circuit,

$\begin{aligned} V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\ \text{or} \quad 120 & =\sqrt{40^2+\left(50-V_C\right)^2} \\ 120^2 & =40^2+\left(50-V_C\right)^2 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad\left(50-V_C\right)^2=12800 \\ & \Rightarrow \quad 50-V_C=80 \sqrt{2} \Rightarrow V_C=10(5-8 \sqrt{2}) \end{aligned}$

We know that, for series R-C circuit

$V^2=V_C^2+V_R^2$

$\begin{aligned} 100 & =64+V_R^2 \\\\ \Rightarrow & V_R^2=36 \Rightarrow V_R=\sqrt{36}=6 \mathrm{~V} \end{aligned}$

$\begin{aligned} & \text { Also, } \tan \phi=\frac{V_C}{V_R} \Rightarrow \tan \phi=\frac{8}{6} \\\\ & \tan \phi=\frac{4}{3} \\\\ & \therefore \phi=\tan ^{-1}\left(\frac{4}{3}\right) \end{aligned}$

Rise of current in $L-R$ circuit is given by

$I=I_0\left(1-e^{-t / \tau}\right)$

where, $\quad I_0=\frac{E}{R}=\frac{5}{5}=1 \mathrm{~A}$

Now, $\tau=\frac{L}{R}=\frac{10}{5}=2 \mathrm{~s}$

After $2 \mathrm{~s}$, i.e. at $t=2 \mathrm{~s}$

Rise of current, $I=\left(1-e^{-1}\right) \mathrm{A}$

$\therefore \tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}$

$\phi$ being the angle by which the current leads the voltage.

Given, $\phi=45^{\circ}$

$\begin{aligned} \therefore \quad & \tan 45^{\circ} =\frac{\omega L-\frac{1}{\omega C}}{R} \Rightarrow 1=\frac{\omega L-\frac{1}{\omega C}}{R} \\ & R =\omega L-\frac{1}{\omega C} \Rightarrow \omega C=\frac{1}{\omega L-R} \\ \Rightarrow \quad & C =\frac{1}{\omega(\omega L-R)}=\frac{1}{2 \pi f(2 \pi f L-R)} \end{aligned}$

For circuit A

Impedence, $Z_A=\sqrt{R^2+\frac{1}{\omega^2 c^2}}$

Current in circuit $I_R^A=\frac{V}{\sqrt{R^2+\frac{1}{\omega^2 c^2}}}$ ..... (i)

Potential difference across $C$,

$V_C^A=I_R^A \times \frac{1}{\omega c}=\frac{V}{\sqrt{(R \omega c)^2+1}}$

For circuit B

Impedence, $Z_B=\sqrt{R^2+\frac{1}{(4 \omega)^2 c^2}}$

Current in circuit, $I_R^B=\frac{V}{\sqrt{R^2+\frac{1}{(4 \omega c)^2}}}$

Potential difference across $C$,

$V_C^B=I_R^B \times \frac{1}{4 \omega c}=\frac{V}{\sqrt{(4 R \omega c)^2+1}}$ ..... (ii)

Hence from above Eqs. (i) and (ii), we conclude that

$I_R^B> I_R^A \text { and } V_C^A> V_C^B .$

The average value of alternating emf $E=E_0 \sin \omega t$ over a positive half cycle is

$E_{\text {av }} \text { half cycle }=\frac{2 E_0}{\pi}=0.6369$

Percentage of $E_{\mathrm{av}}=0.6369 \times 100=63.69 \%$

Also rms value of emf is given as

$E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}=0.7072$

Percentage

$\begin{aligned} E_{\mathrm{rms}} & =0.7072 \times 100 \\ & =70.72 \% \end{aligned}$