Alternating Current

(A) Any static charge produces electric field $\forall$ invariant with time. Uniformly charged dielectric does the same.

(B) Rotating charged behaves like a current. hence, magnetic field is produced around the ring and also if there is current in rotating ring, ring has magnetic moment.

(C) Since, net charge is zero there will be no time independent electric field. The current produces magnetic field and magnetic moment.

(D) A changing magnetic field will be produced. This will create a induced electric field.

Also a changing magnetic moment will be produced.

Charge on capacitor at time $t$ is

$ q=q_0\left(1-e^{-1 / \mathrm{T}}\right) $

Here, $q_0=\mathrm{CV}$ and $d t=2 \mathrm{~T}$

$ \therefore q=\mathrm{CV}\left(1-e^{2 T / T}\right)=\mathrm{CV}\left(1-e^2\right) $

$ \begin{aligned} &\begin{aligned} q & =\mathrm{Q}_0 \cos \omega t \\ i & =\frac{d q}{d t}=\mathrm{Q}_0 \omega \sin \omega t \end{aligned}\\ &\text { From conservation of energy. }\\ &\begin{aligned} \frac{1}{2} \mathrm{LI}_{\max }^2 & =\frac{1}{2} \mathrm{CV}^2 \\ \mathrm{I}_{\max } & =\sqrt{\frac{\mathrm{CV}^2 \times \frac{1}{2}}{\frac{1}{2} \mathrm{~L}}}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{~L}}} \end{aligned} \end{aligned} $

Comparing the LC oscillations with normal SHM,

$ \text { We get } \quad \frac{d^2 \mathrm{Q}}{d t^2}=-\omega^2 \mathrm{Q} $

$ \begin{array}{ll} \Rightarrow & \mathrm{Q}=-\frac{1}{w^2} \frac{d^2 \mathrm{Q}}{d t^2} \\ \text { Here, } & w^2=\frac{1}{\mathrm{LC}} \Rightarrow \frac{1}{w^2}=\mathrm{LC} \\ \therefore & \mathrm{Q}=-\mathrm{LC} \frac{d^2 \mathrm{Q}}{d t^2} \end{array} $

For maximum power, ${X_L} = X{}_C,$ which yields

$C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}$

$\therefore$ $C = 0.1 \times {10^{ - 5}}F = 1\mu F$

Power factor $ = \cos \phi = {R \over Z} = {{12} \over {15}} = {4 \over 5} = 0.8$

The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit:

• In a purely resistive ($R$) circuit, the current and emf are in phase, meaning the phase difference is $0$.


• In a purely inductive ($L$) circuit, the current lags behind the emf by $\frac{\pi}{2}$, meaning the phase difference is $\frac{\pi}{2}$.


• In a purely capacitive ($C$) circuit, the current leads the emf by $\frac{\pi}{2}$, again meaning the phase difference is $\frac{\pi}{2}$.


• In an $L$-$R$ or $L$-$C$ circuit, the phase difference depends on the relative values of $L$, $R$, and $C$ and can be anywhere between $0$ and $\frac{\pi}{2}$.

Therefore, if the phase difference between the alternating current and emf is $\frac{\pi}{2}$, then the circuit cannot contain only a resistor ($R$) since that would give a phase difference of $0$. So, the answer is Option A: $R,L$. The phase difference would not be $\frac{\pi}{2}$ if the circuit contains both a resistor and an inductor.

Since the phase difference between $L$ & $C$ is $\pi ,$

$\therefore$ net voltage difference across $LC=50-50=0$

$D.C.$ ammeter measure average current in $AC$ current, average current is zero for complete cycle. Hence reading will be zero.

For resonant frequency to remain same $LC$ should be const. $LC=$ const

$ \Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}$

Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.

When the capacitor is completely charged, the total energy in the $L.C$ circuit is with the capacitor and that energy is $E = {1 \over 2}{{{Q^2}} \over C}$

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get ${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$ where $Q'$ is the charge on plate of the capacitor

$\therefore$ ${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$

$ \Rightarrow Q' = {Q \over {\sqrt 2 }}$

The impedance triangle for resistance $\left( R \right)$ and inductor $(L)$ connected in series is shown in the figure.



Power factor $\cos \phi = {R \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$

${N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?$

For a transformer ${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$

$ \Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$