Class 12th

Peak voltage is $\sqrt[]{2}$  times rms voltages in ac.

In (a) & (c) circuits, both the junctions are in same biasing conditions so offers equal resistances.

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

Capacitive reactance $=\frac{1}{\omega \mathrm{C}}={\mathrm{X}}_{\mathrm{c}}$ (say) on decreasing the operating frequency $\omega$  reduces

As $X_c$ is inversely proportional to $\omega$ the value of $X_c$  increase

$\therefore {\mathrm{I}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{D}}$

$=\frac{{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{c}}}$

As ${\mathrm{X}}_{\mathrm{c}}$  increases, therefore displacement current Id decreases.

For transformer

$\left(\frac{i_p}{i_s}\right)=\left(\frac{V_s}{V_P}\right)=\left(\frac{N_s}{N_P}\right)$

For bulb, $\mathrm{V}\mathrm{s}=12\mathrm{V}$

$\begin{array}{rr}& {\mathrm{i}}_{\mathrm{s}}=\left(\frac{60}{12}\right)=5\mathrm{A}\\ & \Rightarrow \frac{{\mathrm{i}}_{\mathrm{p}}}{5}=\left(\frac{12}{220}\right)\\ & \mathrm{}{\mathrm{i}}_{\mathrm{P}}=\frac{60}{220}=\left(\frac{3}{11}\right)=0.27\mathrm{A}\end{array}$

For resonance frequency

$\begin{array}{rr}& \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}\\ & \Rightarrow \omega =\frac{1}{\sqrt[]{\mathrm{L}\mathrm{C}}}=\frac{1}{\sqrt[]{10*{10}^{-3}*1*{10}^{-6}}}\\ & =\frac{1}{\sqrt[]{{10}^{-8}}}={10}^4\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}\\ & \mathrm{f}=\left(\frac{\omega }{2\pi }\right)=\frac{{10}^4}{2\pi }=1.59\mathrm{k}\mathrm{H}\mathrm{z}\end{array}$

Magnetic energy stored in an inductor $=\frac{1}{2}{\mathrm{L}\mathrm{I}}^2$

$\begin{array}{rr}& =\frac{1}{2}*4*{10}^{-6}*(2{)}^2\\ & =8\mu \mathrm{J}\end{array}$

${\text{?}\mathrm{}}_{\mathrm{s}}\stackrel{?}{\mathrm{B}}\cdot \mathrm{d}\stackrel{?}{\mathrm{A}}=0\mathrm{}$ (No monopole exist)

For conductors α $\alpha$  is (+ve)

For semiconductors $\alpha$  is (-ve)