NCERT Exemplar · Class 12 Physics Alternating Current

Match the energy terms one by one.\[\tfrac12kx^2\;\leftrightarrow\;\frac{q^2}{2C},\qquad \tfrac12mv^2\;\leftrightarrow\;\tfrac12Li^2 \]\[x\leftrightarrow q,\quad v\leftrightarrow i,\quad m\leftrightarrow L,\quad k\leftrightarrow\frac1C \]\[\omega=\sqrt{\frac{k}{m}}\;\leftrightarrow\;\omega_0=\frac{1}{\sqrt{LC}} \] Answer: The capacitor's electrical energy is analogous to potential energy; the inductor's magnetic energy to kinetic energy.

At very high \(\displaystyle \omega\), capacitors act as wires and inductors as open circuits.\[X_C=\frac{1}{\omega C}\to0,\qquad X_L=\omega L\to\infty \]\(\displaystyle L_1\) and \(\displaystyle L_2\) open, so \(\displaystyle R_2\) carries no current; \(\displaystyle C_1\), \(\displaystyle C_2\) shorted, so the current runs through \(\displaystyle R_1\) and \(\displaystyle R_3\) in series.\[Z_{eff}=R_1+R_3 \] Answer: \(\displaystyle Z_{eff}=R_1+R_3\).

Both circuits have the same source and the same \(\displaystyle R\).\[I_a=\frac{V_{rms}}{R},\qquad I_b=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)^2}} \](a) \[I_b=I_a\;\Rightarrow\;X_L=X_C\;\Rightarrow\;\omega=\frac{1}{\sqrt{LC}} \](b) \[\sqrt{R^2+(X_L-X_C)^2}\ge R\;\Rightarrow\;I_b\le I_a \] Answer: (a) At resonance, \(\displaystyle \omega=1/\sqrt{LC}\). (b) No; \(\displaystyle I_b\le I_a\) always.

\[v=V_m\sin\omega t,\qquad i=I_m\sin(\omega t-\phi) \]\[p=vi=\frac{V_mI_m}{2}\left[\cos\phi-\cos(2\omega t-\phi)\right] \]\(\displaystyle p<0\) whenever \(\displaystyle v\) and \(\displaystyle i\) have opposite signs, which happens for \(\displaystyle \phi\neq0\); the circuit then returns energy to the source.\[P_{av}=V_{rms}I_{rms}\cos\phi,\qquad |\phi|\le90^\circ\;\Rightarrow\;P_{av}\ge0 \] Answer: Yes, \(\displaystyle p\) can be negative (\(\displaystyle \phi\neq0\)); no, \(\displaystyle P_{av}\) cannot, for a passive circuit.

Bandwidth is the width of the curve at \(\displaystyle I_{max}/\sqrt2\).\[I_{max}=1.0\ \mathrm{A}\ \text{ at }\ \omega_0=1.0\ \mathrm{rad\,s^{-1}} \]\[I=\frac{I_{max}}{\sqrt2}=\frac{1.0}{1.414}\approx0.71\ \mathrm{A} \]\[\omega_1\approx0.75\ \mathrm{rad\,s^{-1}},\qquad \omega_2\approx1.25\ \mathrm{rad\,s^{-1}} \]\[\Delta\omega=\omega_2-\omega_1\approx1.25-0.75=0.5\ \mathrm{rad\,s^{-1}} \] Answer: \(\displaystyle \Delta\omega\approx0.5\ \mathrm{rad\,s^{-1}}\), marked between \(\displaystyle \omega_1\) and \(\displaystyle \omega_2\) at \(\displaystyle 0.71\ \mathrm{A}\).

In one period \(\displaystyle 2T\): \(\displaystyle i=+1\ \mathrm{A}\) for a total time \(\displaystyle T\), and \(\displaystyle i=-2\ \mathrm{A}\) for a time \(\displaystyle T\).\[I_{rms}^2=\frac{1}{2T}\int_0^{2T}i^2\,dt=\frac{(1)^2\,T+(-2)^2\,T}{2T} \]\[I_{rms}^2=\frac{1+4}{2}=2.5\ \mathrm{A^2}\;\Rightarrow\;I_{rms}=\sqrt{2.5}\approx1.58\ \mathrm{A} \]Square before averaging: the mean of \(\displaystyle i\) is \(\displaystyle -0.5\ \mathrm{A}\), not \(\displaystyle I_{rms}\). Answer: \(\displaystyle I_{rms}=\sqrt{2.5}\approx1.58\ \mathrm{A}\), drawn as the dashed lines at \(\displaystyle \pm1.58\ \mathrm{A}\).

\[\tan\phi=\frac{X_L-X_C}{R}=\frac{\omega L-\dfrac{1}{\omega C}}{R} \]\[\omega\to0:\quad X_C\gg X_L\;\Rightarrow\;\phi\to-90^\circ \] \[\omega=\omega_0=\frac{1}{\sqrt{LC}}:\quad X_L=X_C\;\Rightarrow\;\phi=0 \] \[\omega\to\infty:\quad X_L\gg X_C\;\Rightarrow\;\phi\to+90^\circ \]\(\displaystyle \omega\uparrow\Rightarrow\phi\) rises from negative, through zero at resonance, to positive. Answer: \(\displaystyle \phi\) changes sign from negative (capacitive) to positive (inductive), passing through zero at \(\displaystyle \omega_0\).