NCERT Exemplar · Class 12 Physics Alternating Current

Lagging current means \(\displaystyle X_L>X_C\), so \(\displaystyle X_C-X_L\) is negative.\[\tan\phi=-\tfrac34 \;\Rightarrow\; \cos\phi=\tfrac45 \]\[P=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi \;\Rightarrow\; I_{\mathrm{rms}}=\frac{2000\ \mathrm{W}}{\sqrt{50000}\ \mathrm{V}\times0.8}=\sqrt{125}\ \mathrm{A}\approx11.2\ \mathrm{A} \](i)\[P=I_{\mathrm{rms}}^{2}R \;\Rightarrow\; R=\frac{2000}{125}=16\ \Omega \](ii)\[\tan\phi=\frac{X_C-X_L}{R} \;\Rightarrow\; X_C-X_L=-\frac34\times16=-12\ \Omega \](iii)\[I_M=\sqrt2\,I_{\mathrm{rms}}=\sqrt{250}\ \mathrm{A}\approx15.8\ \mathrm{A} \]Check:\[Z=\frac{R}{\cos\phi}=\frac{16}{0.8}=20\ \Omega,\qquad \frac{V_{\mathrm{rms}}}{Z}=\frac{\sqrt{50000}}{20}=11.2\ \mathrm{A} \]Second device on the same mains:\[R'=32\ \Omega,\qquad X_C'-X_L'=-24\ \Omega,\qquad Z'=2Z=40\ \Omega \]\[\tan\phi'=\frac{-24}{32}=-\frac34 \quad(\phi\ \text{unchanged}) \]\[I_M'=\frac{\sqrt2\,\sqrt{50000}}{40}=\frac{I_M}{2}\approx7.9\ \mathrm{A},\qquad P'=I_{\mathrm{rms}}'^{\,2}R'=5.59^{2}\times32\approx1000\ \mathrm{W}=\frac P2 \]Answer: \(\displaystyle R=16\ \Omega\), \(\displaystyle X_C-X_L=-12\ \Omega\), \(\displaystyle I_M\approx15.8\ \mathrm{A}\). For the second device \(\displaystyle R\) and \(\displaystyle X_C-X_L\) double, \(\displaystyle \phi\) is unchanged, \(\displaystyle I_M\) halves to about \(\displaystyle 7.9\ \mathrm{A}\) and the power halves to \(\displaystyle 1\ \mathrm{kW}\).

Current flows out and back, so the line is two wires of length \(\displaystyle l=10\ \mathrm{km}\). Caution: the loss uses the line current at the transmission voltage, not the $\displaystyle 220$ V of the town.\[A=\pi r^{2}=\pi\,(0.5\times10^{-2}\ \mathrm{m})^{2}=7.85\times10^{-5}\ \mathrm{m^{2}} \]\[R=\rho\,\frac{2l}{A}=\frac{1.7\times10^{-8}\times2\times10^{4}}{7.85\times10^{-5}}\ \Omega=4.33\ \Omega \](i) Transmission at $\displaystyle 220$ V\[I=\frac{P}{V}=\frac{10^{6}\ \mathrm{W}}{220\ \mathrm{V}}=4545\ \mathrm{A} \]\[P_{\mathrm{loss}}=I^{2}R=(4545)^{2}\times4.33\approx8.9\times10^{7}\ \mathrm{W} \]\[\frac{P_{\mathrm{loss}}}{P}=\frac{8.9\times10^{7}}{10^{6}}\approx89 \]The line would waste about $\displaystyle 89$ times the power sent, so this is not feasible.(ii) Step-up to $\displaystyle 11000$ V\[I=\frac{10^{6}\ \mathrm{W}}{11000\ \mathrm{V}}=90.9\ \mathrm{A} \]\[P_{\mathrm{loss}}=I^{2}R=(90.9)^{2}\times4.33\approx3.6\times10^{4}\ \mathrm{W} \]\[\frac{P_{\mathrm{loss}}}{P}=\frac{3.6\times10^{4}}{10^{6}}\approx0.036=3.6\% \]Answer: (i) loss \(\displaystyle \approx89\) times the power sent (\(\displaystyle \approx8.9\times10^{7}\ \mathrm{W}\)), so not feasible; (ii) loss \(\displaystyle \approx3.6\%\) of the power sent.

\(\displaystyle R\) and the \(\displaystyle L\)-\(\displaystyle C\) branch are in parallel across the source, so each sees \(\displaystyle v=v_m\sin\omega t\), and \(\displaystyle i=i_1+i_2\). Caution: \(\displaystyle i_1\) and \(\displaystyle i_2\) are \(\displaystyle 90^\circ\) apart, so add them as phasors, not as plain numbers.\[i_1=\frac{v_m}{R}\sin\omega t \quad(\text{in phase with }v) \]\[i_2=\frac{v_m}{X_C-X_L}\sin\!\left(\omega t+\frac\pi2\right)=\frac{v_m}{X_C-X_L}\cos\omega t,\qquad X_C=\frac1{\omega C},\ X_L=\omega L \]\[i=i_1+i_2=\frac{v_m}{R}\sin\omega t+\frac{v_m}{X_C-X_L}\cos\omega t=i_m\sin(\omega t+\phi) \]\[i_m=v_m\sqrt{\frac1{R^{2}}+\frac1{(X_C-X_L)^{2}}},\qquad \tan\phi=\frac{R}{X_C-X_L} \]Define \(\displaystyle Z\) by\[\frac1Z=\sqrt{\frac1{R^{2}}+\frac1{(X_C-X_L)^{2}}} \;\Rightarrow\; i_m=\frac{v_m}{Z},\qquad i_{\mathrm{rms}}=\frac{v_{\mathrm{rms}}}{Z} \]\[Z=\frac{R\,\lvert X_C-X_L\rvert}{\sqrt{R^{2}+(X_C-X_L)^{2}}}=\left[\frac1{R^{2}}+\frac{\omega^{2}C^{2}}{(1-\omega^{2}LC)^{2}}\right]^{-1/2} \]If \(\displaystyle X_C<X_L\), \(\displaystyle i_2\) lags \(\displaystyle v\) and \(\displaystyle \phi<0\); the same formulas hold.Answer: \(\displaystyle i=i_m\sin(\omega t+\phi)\) with \(\displaystyle \tan\phi=\dfrac{R}{X_C-X_L}\), \(\displaystyle i_m=\dfrac{v_m}{Z}\), and \(\displaystyle \dfrac1Z=\sqrt{\dfrac1{R^{2}}+\dfrac1{(X_C-X_L)^{2}}}\).

(i) Multiply by \(\displaystyle i=\dfrac{dq}{dt}\), and use \(\displaystyle Li\dfrac{di}{dt}=\dfrac{d}{dt}\!\left(\tfrac12Li^{2}\right)\), \(\displaystyle \dfrac qC\,i=\dfrac{d}{dt}\!\left(\dfrac{q^{2}}{2C}\right)\):\[Li\frac{di}{dt}+Ri^{2}+\frac qC\,i=v_m i\sin\omega t \]\[\frac{d}{dt}\!\left(\tfrac12Li^{2}\right)+Ri^{2}+\frac{d}{dt}\!\left(\frac{q^{2}}{2C}\right)=vi \](ii) Term by term:\[\frac{d}{dt}\!\left(\tfrac12Li^{2}\right):\ \text{rate of change of energy in the inductor's magnetic field} \]\[Ri^{2}:\ \text{rate of Joule heating in } R \]\[\frac{d}{dt}\!\left(\frac{q^{2}}{2C}\right):\ \text{rate of change of energy in the capacitor's electric field} \]\[vi:\ \text{power supplied by the source} \](iii) With \(\displaystyle U_L=\tfrac12Li^{2}\) and \(\displaystyle U_C=\dfrac{q^{2}}{2C}\):\[vi=\frac{d}{dt}\left(U_L+U_C\right)+Ri^{2} \]Source power = rate of storing energy in \(\displaystyle L\) and \(\displaystyle C\) + rate of dissipation in \(\displaystyle R\).(iv) Over one period \(\displaystyle T=\dfrac{2\pi}{\omega}\), \(\displaystyle i\) and \(\displaystyle q\) return to their starting values, so\[\int_0^T\frac{d}{dt}\!\left(\tfrac12Li^{2}\right)dt=0,\qquad \int_0^T\frac{d}{dt}\!\left(\frac{q^{2}}{2C}\right)dt=0 \;\Rightarrow\; \int_0^T vi\,dt=R\int_0^T i^{2}\,dt \]Put \(\displaystyle v=v_m\sin\omega t\), \(\displaystyle i=i_m\sin(\omega t+\phi)\):\[\frac{v_mi_mT}{2}\cos\phi=\frac{Ri_m^{2}T}{2}\;\Rightarrow\;\cos\phi=\frac{Ri_m}{v_m}>0 \]\[\cos\phi>0\;\Rightarrow\;-\frac\pi2<\phi<\frac\pi2 \]Answer: \(\displaystyle vi=\dfrac{d}{dt}\left(\tfrac12Li^{2}+\dfrac{q^{2}}{2C}\right)+Ri^{2}\); over a cycle \(\displaystyle \cos\phi=Ri_m/v_m>0\), so the phase difference between \(\displaystyle v\) and \(\displaystyle i\) is acute.

(i) Loop rule with \(\displaystyle i=\dfrac{dq}{dt}\):\[L\frac{d^{2}q}{dt^{2}}+R\frac{dq}{dt}+\frac qC=v_m\sin\omega t \](ii) After \(\displaystyle t_o\) only \(\displaystyle L\) and \(\displaystyle C\) remain. Caution: inductor current and capacitor charge cannot jump, so the values at \(\displaystyle t_o\) carry over. With \(\displaystyle i_o=\left.\dfrac{dq}{dt}\right|_{t_o}\) and \(\displaystyle q_o=q(t_o)\):\[U_L=\tfrac12Li_o^{2},\qquad U_C=\frac{q_o^{2}}{2C} \]In steady state, \(\displaystyle i=i_m\sin(\omega t+\phi)\) and \(\displaystyle q=-\dfrac{i_m}{\omega}\cos(\omega t+\phi)\), so\[U_L=\tfrac12Li_m^{2}\sin^{2}(\omega t_o+\phi),\qquad U_C=\frac{i_m^{2}}{2\omega^{2}C}\cos^{2}(\omega t_o+\phi) \]\[i_m=\frac{v_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}},\qquad \tan\phi=\frac{X_C-X_L}{R} \](iii) With \(\displaystyle t'=t-t_o\), no source and no resistance:\[L\frac{d^{2}q}{dt^{2}}+\frac qC=0 \;\Rightarrow\; q=q_o\cos\omega_0t'+\frac{i_o}{\omega_0}\sin\omega_0t',\qquad \omega_0=\frac1{\sqrt{LC}} \]The charge sloshes back and forth through \(\displaystyle L\) without damping, and the current reverses every half cycle. Energy passes between \(\displaystyle C\) (all of it when \(\displaystyle i=0\)) and \(\displaystyle L\) (all of it when \(\displaystyle q=0\)):\[U_L+U_C=\tfrac12Li_o^{2}+\frac{q_o^{2}}{2C}=\frac{q_m^{2}}{2C}=\text{constant},\qquad q_m^{2}=q_o^{2}+LCi_o^{2} \]Answer: (i) \(\displaystyle L\ddot q+R\dot q+q/C=v_m\sin\omega t\); (ii) \(\displaystyle U_L=\tfrac12Li_o^{2}\), \(\displaystyle U_C=q_o^{2}/2C\); (iii) undamped \(\displaystyle LC\) oscillation at \(\displaystyle \omega_0=1/\sqrt{LC}\) with constant total energy.