Physics M.C.Q (1 Marks)

$\because \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R} \times 500}{\mathrm{R}+500} \Rightarrow \mathrm{By} \mathrm{\,eq}^{\mathrm{n}}$   $(1)$

$\Rightarrow 10 \times 10^{-6}=\frac{4 \times 10^{-3}}{\mathrm{R}(500)} \times(\mathrm{R}+500)$

$\Rightarrow \mathrm{R}=2000 \Omega=2 \mathrm{K} \Omega$

$\Rightarrow \mathrm{V}^{2}=\frac{\mathrm{Li}^{2}}{\mathrm{C}}$

$=\frac{2 \times 2^{2}}{4 \times 10^{-6}}=2 \times 10^{6}$

$\Rightarrow \mathrm{V}=\sqrt{2} \times 10^{3} \mathrm{\,volt}$

$\mathrm{V}$ is of order $10^{3} \mathrm{\,V}$

$e=L \frac{d i}{d t}=L \frac{E}{R} e^{-t R / L} \frac{R}{L}$ curve $C$ represents $e$

since ${\rm{I}} = {{\rm{I}}_0}{{\rm{e}}^{ - {{\rm{t}}_0}/\tau }} \Rightarrow \frac{{{{\rm{I}}_0}}}{\alpha } = {{\rm{I}}_0}{{\rm{e}}^{ - {{\rm{t}}_0}/\tau }}$

$\alpha=e^{t_{0} / \tau} \Rightarrow t_{0}=\tau \log _{e} \alpha$

$\boxed{\tau  = \frac{{{t_0}}}{{{{\log }_e}\alpha }}}$

$\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}$

$\therefore$ $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{3 \mathrm{mH}}{3 \Omega}=1 \mathrm{\,ms}$

Long after $\mathrm{S}$ is closed $= \mathrm{i}=\frac{10}{5}=2 \mathrm{\,amp}$

difference $=1$ $\mathrm{amp}$

At $t=0$, inductor offers infinite resistance

So, $I_1=\frac{E}{R_1}, I_2=I_3=0$

At $t=0, L$ offers infinite resistance

At $t \rightarrow \infty, C$ offers infinite resistance

So, in both the cases, $l=\frac{\varepsilon}{2 R}$

Hence ratio is $1: 1$.

$i=\frac{E}{R_2}\left[1-e^{-R_2 t / L}\right]$

$\Rightarrow \frac{d i}{d t}=\frac{E}{R_2} \cdot \frac{R_2}{L} \cdot e^{-R_2 t / L}=\frac{E}{L} e^{-\frac{R_2 t}{L}}$

Hence, potential drop across

$L =\left(\frac{E}{L} e^{-R_2 t / L}\right) L=12 e ^{-5 t }\,V$

$E=\frac{R}{2} \frac{d B}{d t}$

So, on electron at $\mathrm{P}(\mathrm{F})=\mathrm{e} \mathrm{E}$

$\mathrm{F}=\frac{\mathrm{eR}}{2} \frac{\mathrm{dB}}{\mathrm{dt}}$

$\therefore $ acceleration of $\mathrm{e}^{-}$ at $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} \mathrm{R}}{2 \mathrm{m}} \frac{\mathrm{dB}}{\mathrm{dt}}$

(opposite in direction of induced current)

$\mathrm{E}=\frac{\mathrm{R}}{2} \frac{\mathrm{dB}}{\mathrm{dt}} \Rightarrow=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{eR}}{2 \mathrm{m}} \frac{\mathrm{dB}}{\mathrm{dt}}$

$\mathrm{E}(2 \pi r)=\pi r^{2} \mu_{0} \mathrm{nI}_{0} \omega \cos \omega t$

$\mathrm{E}=\left(\frac{\mu_{0} \mathrm{n} \mathrm{I}_{0} \mathrm{r} \omega}{2}\right) \cos \omega \mathrm{t}$

$\mathrm{E} 2 \pi \mathrm{r}=\pi \mathrm{a}^{2} \frac{\mathrm{d} \mathrm{B}}{\mathrm{dt}}$

$\mathrm{E}=\frac{\mathrm{a}^{2} \frac{\mathrm{d} \mathrm{B}}{\mathrm{dt}}}{\mathrm{r}}$

$E=\frac{a^{2} d B / d t}{2 r}=\frac{\left(5 \times 10^{-2}\right)^{2} \times 4}{2 \times 10 \times 10^{-2}}$

$\mathrm{E}=\frac{25 \times 10^{-4} \times 4}{2 \times 10^{-1}}=\frac{100}{2} \times 10^{3}=\frac{10^{-1}}{2}=0.05 \mathrm{\,V} / \mathrm{m}$

Induced emf $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})$

$=-A \frac{d B}{d t}=-\frac{1}{2}\left(1^{2}\right) \times(-2)$

$\Rightarrow e=1 \,V$

Since magnetic field $( \times )$ decreasing so according to Lenz's law direction of induced current in upper part of square will be clockwise i.e. from $\mathrm{A}$ to $\mathrm{C}$ or in other words emf induces in a direction opposite to the main emf so resultant emf $=10-1=9 \,V.$

Rate of increase of $B=\frac{d B}{d t}$

$E_{ m } \times 2 \pi r=-\frac{d \phi}{d t}$

or $E \times 2 \pi r=-A \frac{d B}{d t}$

$E \times 2 \pi r=-\pi r^2 \frac{d B}{d t}$

$\therefore E=\frac{-r}{2} \frac{d B}{d t}$

$\Rightarrow \frac{{200}}{{100}} = \frac{{{V_s}}}{{120}} $

$\Rightarrow {V_s} = 240\;V$

also $\frac{{{V_s}}}{{{V_p}}} = \frac{{{i_p}}}{{{i_s}}}$

$\Rightarrow \frac{{240}}{{120}} = \frac{{10}}{{{i_s}}}$

$\Rightarrow {i_s} = 5\;A$

$\Rightarrow \frac{1}{{20}} = \frac{{{V_s}}}{{2400}}$

$\Rightarrow {V_s} = 120\;V$ 

For $100\%$ efficiency ${V_s}{i_s} = {V_p}{i_p}$ 

$ \Rightarrow 120 \times 80 = 2400\;{i_p}$

$\Rightarrow {i_p} = 4\;A$

$\Rightarrow {V_p} = \frac{{200}}{5} = 40\;V$

Also ${i_p}{V_p} = {i_s}{V_s}$

$\Rightarrow {i_p} = {i_s}\frac{{{V_s}}}{{{V_p}}} = 8 \times 5 = 40\;A$

$\Rightarrow \frac{{250}}{{100}} = \frac{{{V_s}}}{{28/\sqrt 2 }}$

$\Rightarrow {V_s} = 50\;V$

${V_s}{i_s} = {V_p}{i_p}$

$\Rightarrow \frac{{{V_s}}}{{{V_p}}} = \frac{{{i_p}}}{{{i_s}}} = \frac{{{N_s}}}{{{N_p}}}$

$ \Rightarrow \frac{{{i_p}}}{4} = \frac{{25}}{{100}}$

$\Rightarrow {i_p} = 1\;A$

${P_{out}} = {V_s}{i_s} \Rightarrow 30 = 6 \times {i_s}$

$\Rightarrow {i_s} = 5\;A$

From $\frac{{{V_s}}}{{{V_p}}} = \frac{{{i_p}}}{{{i_s}}}$

$\Rightarrow \frac{6}{{200}} = \frac{{{i_p}}}{5}$

$\Rightarrow {i_p} = 0.15\;A$

$\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}$

$\Rightarrow \frac{{100}}{{{V_p}}} = \frac{{10}}{1} $

$\Rightarrow {V_p} = 10\;V$ 

$\frac{{{i_p}}}{{{i_s}}} = \frac{{{N_s}}}{{{N_p}}}$

$\Rightarrow \frac{{{i_p}}}{{0.5}} = \frac{{10}}{1},\;{i_p} = 5\;amp$

$\frac{{{V_p}}}{{{V_s}}} = \frac{{{N_p}}}{{{N_s}}} \Rightarrow \frac{{(1000/8)}}{{500}} = \frac{{100}}{{{N_s}}} \Rightarrow {N_s} = 400$

==> $\frac{{80}}{{100}} = \frac{{200 \times {I_s}}}{{4 \times {{10}^3}}}$ 

==> ${I_S} = \frac{{80}}{{100}} \times \frac{{4 \times 1000}}{{200}} = 16\,A$ 

Also, ${E_p}{I_p} = 4KW$

==>${I_p} = \frac{{4 \times {{10}^3}}}{{100}} = 40\,A$

==> $\frac{{80}}{{100}} = \frac{{20 \times 20}}{{1000 \times {i_l}}}$

==> ${i_l} = \frac{{20 \times 120 \times 100}}{{1000 \times 80}} = 3\,A$.

 

$\mathrm{P}=\mathrm{VI} \cos \theta$

$\cos \theta=1$

$\mathrm{P}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{Z}_{1}}$

$\frac{V_{1}^{2}}{V_{2}^{2}}=\frac{Z_{1}}{Z_{2}}$

$\frac{V_{1}}{V_{2}}=\frac{N_{1}}{N_{2}}$

[By transformer (mutual induction) ]

$\mathrm{V}_{2}=8 \times 120=960 \mathrm{\,volt}$

$\mathrm{I}=\frac{960}{10^{4}}=96 \mathrm{\,mA}$

$\mathrm{N}_{\mathrm{s}}=280^{\circ}$

From the formula

$\frac{\mathrm{i}_{\mathrm{p}}}{\mathrm{i}_{\mathrm{s}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}} \quad$ or $\quad \frac{4}{\mathrm{i}_{\mathrm{s}}}=\frac{280}{140}$

So, $\quad i_{s}=2 \,A$

$\frac{V_{s}}{V_{p}}=\frac{N_{2}}{N_{1}} \Rightarrow V_{s}=\frac{N_{s}}{N_{p}} \times V_{p}$

$\mathrm{V}_{\mathrm{s}}=5 \times 100 \sqrt{2}=500 \sqrt{2} \mathrm{\,V}$

$\mathrm{E}_{\mathrm{S}}=\frac{\mathrm{n}_{\mathrm{S}}}{\mathrm{n}_{\mathrm{P}}} \times \mathrm{E}_{\mathrm{P}}=\frac{200}{100} \times 120=240 \mathrm{\,V}$

$I_{s}=I_{P} \times \frac{n_{P}}{n_{S}}=10 \times \frac{100}{200}=5 \,A$

Also $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{200}=\frac{\pi}{100}$

$\Rightarrow|\mathrm{e}|=\frac{600}{\pi}=190.9 \mathrm{V} \simeq 191 \mathrm{\,V}$

$\frac{140}{280}=\frac{I_{2}}{4} \Rightarrow I_{2}=2 \,A$

$P_{i}=V I=240 \times 0.7=168$ $\mathrm{watt}$

$P_{0}=140$ $\mathrm{watt}$

$1000=500 I$

$(I)$ secondary $=2 \mathrm{A}$

Now $, \frac{I_{1}}{I_{2}}=\frac{N_{2}}{N_{1}} \Rightarrow \frac{8}{2}=\frac{N_{2}}{100}$

$\Rightarrow N_{2}=400$

$\eta=\frac{\text { output power }}{\text { input power }}=\frac{\text { input power }-\text { resistance loss }}{\text { input power }}$$=\frac{ W -I^{2} R }{ V }$

$\eta=\frac{(50 \times 12)-12^{2} R}{50 \times 12}$

$\therefore \frac{30}{100}=\frac{600-144 R}{600}$

$\therefore 0.3 \times 600=600-144 R$

$\therefore 180=600-144 R$

$\therefore R=\frac{600-180}{144}=2.92 \Omega$

$\mathrm{N}_{\mathrm{P}}=6000, \eta=60 \% ; \mathrm{P}_{\mathrm{O}}=9 \mathrm{\,kW}=9 \times 10^{3} \mathrm{\,W}$

Efficiency, $\eta  = \frac{{{\rm{ Output power }}}}{{{\rm{ Input power }}}} = \frac{{{{\rm{P}}_O}}}{{{{\rm{P}}_{\rm{i}}}}}$

$\therefore \mathrm{P}_{i}=\frac{\mathrm{P}_{\mathrm{O}}}{\eta}=\frac{9 \times 10^{3}}{60 / 100}=1.5 \times 10^{4}=15 \mathrm{\,kW}$

$\Rightarrow \mathrm{e}_{\mathrm{s}}=8 \times 120 \mathrm{\,V} \Rightarrow \mathrm{i}_{\mathrm{5}}=\frac{8 \times 120}{10^{4}}=96 \times 10^{-3} \mathrm{\,A}$

$V_{s} i_{s}=V_{p} i_{p} \Rightarrow \frac{V_{s}}{V_{p}}=\frac{i_{p}}{i_{s}}=\frac{N_{s}}{N_{p}} \Rightarrow \frac{i_{p}}{4}=\frac{25}{100} i_{p}=1 A$

Efficiency$=\frac{\text {Output} \text { power of the transformer}}{\text {Input power of the transformer}}=\frac{P_{s}}{P_{p}} \times 100$

In this case, the output power is calculated as follows.

The output voltage is $11 \mathrm{v}$ and current is $90 \mathrm{A}$

Power $=V I=11 \times 90=990 \mathrm{W}$

The input power is calculated as follows.

The input voltage is $220 \mathrm{V}$ and current is $5 \mathrm{A}$

Power $=V I=220 \times 5=1100 \mathrm{W}$

so, $\eta=\frac{990}{1100} \times 100=90 \%$

The efficiency of the transformer is $90 \%$

$e \propto \frac{{di}}{{dt}}$

Here $di/dt$ represent the slope of $i-t$ curve. In the given $i-t$ graph, during the first half time the slope is constant and has positive value and in the next half time, the slope is again constant but has negative value. Hence the correct representation of the curve in $e-t$ graph is $(C)$.