Physics M.C.Q (1 Marks)

$=\frac{1000}{3.14}$

$V_{\mathrm{ms}}=210 \mathrm{~V}$

$i_{\mathrm{ms}}=\frac{V_{\mathrm{ms}}}{X_C}=\frac{210}{X_C}$

$\text { Peak current }=\sqrt{2} i_{\text {mms }}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932$

$=0.93 \mathrm{~A}$

$X_C =\frac{1}{2 \pi \times 50 \times \frac{10^3}{\pi} \times 10^{-6}}=10\,\Omega$

$Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$

$=\sqrt{(10)^2+(5)^2}$

$=5 \sqrt{5}\,\Omega$

$C =1 \times 10^{-6}\,F$

$R =100\,\Omega$

At resonance $X_L=X_C$

$\omega L =\frac{1}{\omega C }$

$f =\frac{1}{2 \pi \sqrt{ LC }}=\frac{1}{2 \pi \sqrt{10 \times 10^{-3} \times 10^{-6}}}=1.59\,KHz$

$X_1=100 \pi \times 2 \times 10^3$ $\quad X_L=\omega L$

$X_1=0.2\,\pi\,\Omega$ $\quad X_2=0$

$X_1=0.628\,\Omega$

$I _{ ms } \times 200=100$

$I _{ ms }=\frac{1}{2}$

So, $I _{\text {peak }}= I _{ rms } \sqrt{2}$

$=\frac{1}{2} \times \sqrt{2}$

$=\frac{1}{\sqrt{2}}=0.707\,A$

$v=\frac{\omega}{2 \pi}=\frac{100}{2 \pi}=\frac{50}{\pi} Hz$

Resonance frequency

$v_{0}=\frac{1}{2 \pi \sqrt{ LC }}=\frac{1}{2 \pi} \sqrt{\frac{1}{10 \times 10 \times 10^{-6}}}$

$=\frac{50}{\pi}\,Hz$

$\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=50\, \mathrm{rad} / \mathrm{sec}$

$\omega_{\min }=\omega_{0}-\frac{\Delta \omega}{2}=46\, \mathrm{rad} / \mathrm{sec}$

$\omega_{\max }=\omega_{0}-\frac{\Delta \omega}{2}=54 \,\mathrm{rad} / \mathrm{sec}$

$\mathrm{I}_{\mathrm{RMS}}=\frac{\mathrm{I}}{\sqrt{2}}=10 \,\mathrm{~A}$

$V_{\mathrm{RMS}}=\sqrt{V_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$

$=\sqrt{(40)^{2}+(40-10)^{2}}$

$=50\, \mathrm{~V}$

$Z=\frac{\mathrm{V}_{\mathrm{RMS}}}{\mathrm{I}_{\mathrm{RM}}}=\frac{50 \mathrm{~V}}{10 \mathrm{~V}}=5 \,\Omega$

Hence impedance $z=\sqrt{R^{2}+(\omega L)^{2}},$ will increase,

so from $i=\frac{v}{z}$ current will decrease and from $P =\frac{ V _{0}^{2} R }{2 z ^{2}}$ power of bulb will decrease.

$=200 \times 40 \times 10^{-6} \times 2 \pi \times 50$

$=2.5 A$

When $L$ removed tan $\phi=\frac{ X _{ L }}{ R }$

$\frac{ X _{ C }}{ R }=\frac{ X _{ L }}{ R } \Rightarrow$ Resonance

$Z = R$

$\cos \phi=\frac{ R }{ Z }=\frac{ R }{ R }=1$

In D.C Source Capacitor a would provide Infinite resistance but current is present in the given circuit. it means resistor and Inductor are in circuit

$Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}} ;$ where $X_{C}=$ capacitive reactance and $X_{L}=$ inductive reactance.

Also $X_{C}=\frac{1}{\omega C}$ and $X_{L}=\omega L$

$\therefore \quad z=\sqrt{(50)^{2}+\left(\frac{1}{314 \times 100 \times 10^{-6}}-314 \times 20 \times 10^{-3}\right)^{2}}$

or $\quad Z=56\, \Omega$

The power loss in the circuit is $P_{a v}=\left(\frac{V_{r m s}}{Z}\right)^{2} R$

$\therefore \quad P_{a v}=\left(\frac{10}{(\sqrt{2}) 56}\right)^{2} \times 50=0.79\, \mathrm{W}$

Equivalent circuit of the given circuit

Current drawn from battery,

$i=\frac{\varepsilon}{(R / 2)}=\frac{2 \varepsilon}{R}=\frac{2 \times 18}{9}=4 \,\mathrm{A}$

${Q=\frac{1}{R} \sqrt{\frac{L}{C}}} $

${Q_{1}=\frac{1}{15} \sqrt{\frac{3.5}{30 \times 10^{-6}}}=\frac{100}{15} \sqrt{\frac{35}{3}}=22.77} $

${Q_{2}=\frac{1}{25} \times \sqrt{\frac{1.5}{45 \times 10^{-6}}}=40 \times \sqrt{\frac{5}{90}}=9.43} $

${Q_{3}=\frac{1}{20} \sqrt{\frac{1.5}{35 \times 10^{-6}}}=50 \times \sqrt{\frac{3}{70}}=10.35} $

${Q_{4}=\frac{1}{25} \times \sqrt{\frac{2.5}{45 \times 10^{-6}}}=\frac{40}{\sqrt{30}}=7.30}$

Clearly $Q_{1}$ is maximum of $Q_{1}, Q_{2}, Q_{3},$ and $Q_{4}$.

Hence, option $(a)$ should be selected for better tuning of an $L-C-R$ circuit.

Power factor, $\cos \phi=\frac{R}{Z}$

${=\frac{V_{R}}{V}=\frac{V_{R}}{\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}}}$

${=\frac{80}{\sqrt{(80)^{2}+(100-40)^{2}}}=\frac{80}{100}=0.8}$

$C=50\, \mu \mathrm{F}=50 \times 10^{-6}\, \mathrm{F}$

$R=40\, \Omega, V=10\, \sin \,340\, t=V_{0} \sin \omega t$

$\omega=340 \,\mathrm{rad} \,\mathrm{s}^{-1}, V_{0}=10\, \mathrm{V}$

$X_{L}=\omega L=340 \times 20 \times 10^{-3}=6.8\, \Omega$

$X_{C}=\frac{1}{\omega C}=\frac{1}{340 \times 50 \times 10^{-6}}=\frac{10^{4}}{34 \times 5}=58.82 \,\Omega$

$Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}=\sqrt{(40)^{2}+(58.82-6.8)^{2}}$

$=\sqrt{(40)^{2}+(52.02)^{2}}=65.62\, \Omega$

The peak current in the circuit is

$I_{0}=\frac{V_{0}}{Z}=\frac{10}{65.62} \,\mathrm{A}, \cos \phi=\frac{R}{Z}=\left(\frac{40}{65.62}\right)$

Power loss in $A.C.$ circuit,

${=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi=\frac{1}{2} V_{0} I_{0} \cos \phi}$

${=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}=0.46 \,\mathrm{W}}$

$\mathrm{i}=5 \sin \left(\omega \mathrm{t}+\frac{\pi}{4}\right)$

$\Delta p=\frac{\pi}{4}-\frac{\pi}{6}=\frac{3 \pi-2 \pi}{12}=\frac{\pi}{12}$

$ < {\rm{V}} >  = \sqrt {\frac{{\int_0^{{\rm{T}}/4} {{\rm{V}}_0^2} {\rm{dt}} + \int_{{\rm{T}}/4}^{\rm{T}} {(0)} {\rm{dt}}}}{{\int_0^{\rm{T}} {dt} }}} $

$=\sqrt{\frac{V_{0}^{2}\left(\frac{T}{4}\right)}{T}}=\sqrt{\frac{V_{0}^{2}}{4}}=\frac{V_{0}}{2}$

$0.5 \times 10+V_{C}=12$

$\mathrm{q}=\mathrm{CV}_{\mathrm{C}}=5 \times 7=35\, \mu \mathrm{C}$

$\mathrm{i}_{\mathrm{rms}}=\frac{i_{0}}{\sqrt{2}}$

$\mathrm{i}_{\mathrm{rms}}=30 \mathrm{\,mA}$

${\rm{V}} = 50\sin 200\,\pi \,{\rm{t}}$

Here $\mathrm{V}_{0}=50 \quad $ and $\omega  = 200\,\pi \,{\rm{f}} = 100\,{\mkern 1mu} {\rm{Hz}}$

$=\sqrt{\frac{\mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}}{2}}$

where $\omega=\frac{2 \pi}{\mathrm{T}}$

$<\mathrm{I}^{2}>=\int_{2}^{4} \mathrm{I}^{2} \mathrm{dt}=\int_{2}^{4} 4 \mathrm{t} \mathrm{dt}=\frac{4\left(4^{2}-2^{2}\right)}{2 \times 2}=12$

$\mathrm{I}_{\mathrm{rms}}=\sqrt{<\mathrm{I}^{2}>}=\sqrt{12}=2 \sqrt{3} \mathrm{\,A}$

$=50 \sin (200 \pi \mathrm{t})$

$\mathrm{V}_{\mathrm{O}}=50$ and $\omega=200 \pi$

$\mathrm{f}=\frac{\omega}{2 \pi}=\frac{200 \pi}{2 \pi}=100 \mathrm{\,Hz}$

$\therefore \mathrm{q}=\mathrm{q}_{\mathrm{o}} \cos (\omega \mathrm{t})$

$=24 \cos \left(4 \pi \times \frac{1}{12}\right)$

$=24 \cos \left(\frac{\pi}{3}\right)$

$=24 \times \frac{1}{2}=12$

$\mathrm{i}=2 \sin 100 \pi \mathrm{t}+2 \cos \left(100 \pi \mathrm{t}+30^{\circ}\right)$

$\Rightarrow \mathrm{i}=2 \sin 100 \pi \mathrm{t}+2 \cos 100 \pi \mathrm{tcos} 30^{\circ}$

$-2 \sin 100 \pi \operatorname{tsin} 30^{\circ}$

$\Rightarrow \mathrm{i}=\sin 100 \pi \mathrm{t}+\sqrt{3} \cos 100 \pi \mathrm{t}$

$\Rightarrow \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin (100 \pi \mathrm{t}+\phi)$

Where $I_{\operatorname{m}}=\sqrt{1+(\sqrt{3})^{2}}=2 A$

$I_{\operatorname{rms}}=\frac{I_{m}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} A$

${{\rm{Z}} = \frac{{\rm{V}}}{{{{\rm{I}}_2}}}}$          (for  $AC$)

$Z=\sqrt{R^{2}+X_{L}^{2}}$

$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{fL}$