Alternating Current

Phase difference between I₂ and V, i.e. C $-$ R₂ circuit is given by

$\tan \phi = {{{X_C}} \over {{R_2}}} \Rightarrow \tan \phi = {1 \over {C\omega {R_2}}}$

Substituting the given values, we get

$\tan \phi = {1 \over {{{\sqrt 3 } \over 2} \times {{10}^{ - 6}} \times 100 \times 20}} = {{{{10}^3}} \over {\sqrt 3 }}$

$\therefore$ $\phi$₁ is nearly 90$^\circ$.

Phase difference between I₁ and V, i.e. in L $-$ R₁ circuit is given by

$\tan {\phi _2} = - {{{X_L}} \over {{R_1}}} = - {{L\omega } \over R}$

Substituting the given values, we get

$\tan {\phi _2} = - {{{{\sqrt 3 } \over {10}} \times 100} \over {10}} = - \sqrt 3 $

As, $\tan {\phi _2} = - \sqrt 3 $

$\therefore$ ${\phi _2} = 120^\circ $

Now, phase difference between I₁ and I₂ is

$\Delta \phi = {\phi _2} - {\phi _1} = 120^\circ - 90^\circ = 30^\circ $

It is given that e₀ = 283 V; $\omega$ = 320.

The inductor reactance is X_L = 320 $\times$ 25 $\times$ 10^$-$3 = 8 $\Omega$

The capacitor reactance is

${X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega $

It is given that R = 5 $\Omega$. Therefore, the total impedance is

$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega $

and the phase difference between the voltage across the source and the current is

$\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ $

From the circuit, we have

$R = {{80} \over {10}} = 8\,\Omega $

$10 = {{220} \over {\sqrt {{R^2} + X_L^2} }} \Rightarrow \sqrt {64 + X_L^2} = 22$

$ \Rightarrow X_L^2 = 484 - 64 = 420$

$ \Rightarrow {X_L} = \sqrt {420} = 20.5\,\Omega $

$ \Rightarrow \omega L = 20.5$

$L = {{20.5} \over {2\pi \times 50}} = {{20.5} \over {314}} = 0.065\,H$

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.

$ \begin{aligned} & \Rightarrow i_0=\frac{v_0}{R}=\frac{300}{30}=10 \mathrm{~A} \\ & \text { and } \phi=0 \\ & \begin{array}{l} So (P) \rightarrow 3 \end{array} \end{aligned} $

$ \begin{aligned} z & =\sqrt{R^2+(\omega L)^2} \quad\left(\text { as } x_L=\omega L\right) \\ & =\sqrt{(30)^2+\left(400 \times 100 \times 10^{-3}\right)^2} \\ & =\sqrt{900+1600}=\sqrt{2500} \\ z=50 \Omega & \end{aligned} $

$ \begin{aligned} \text { So } & i_0=\frac{V_0}{2}=\frac{300}{50}=6 \mathrm{~A} \\ \phi & =\tan ^{-1}\left(\frac{x_L}{R}\right)=\tan ^{-1}\left(\frac{\omega L}{R}\right) \\ & =\tan ^{-1}\left(\frac{400 \times 106 \times 10^{-3}}{36}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \\ \Rightarrow \quad \phi & =53^{\circ}(l \mathrm{ag}) \\ & \therefore Q \rightarrow 5 \end{aligned} $

$ \begin{aligned} z & =\sqrt{R^2+\left(x_C-x_L\right)^2} \\ & =\sqrt{R^2+\left(\frac{1}{\omega c}-\omega L\right)^2} \\ & =\sqrt{(30)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 25 \times 10^{-3}\right)^2} \\ z & =\sqrt{900+\left(\frac{1}{20 \times 10^{-3}}-10\right)^2} \end{aligned} $

$ \begin{aligned} &\begin{array}{r} z=\sqrt{900+(50-10)^2}=\sqrt{900+1600} \\ z=\sqrt{2500}=50 \Omega \\ i_0=\frac{v_0}{2}=\frac{300 }{50}=6 \mathrm{~A} \\ \phi=\tan ^{-1}\left[\frac{x_L-x_C}{R}\right]=\tan ^{-1}\left[\frac{10-50}{30}\right] \\ =\tan ^{-1}\left|\frac{40 }{30}\right|=53^{\circ} \text { (Lead) } \end{array}\\ &(R) \rightarrow 2 \end{aligned} $

$ \begin{aligned} z & =\sqrt{(60)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 125 \times 10^{-3}\right)^2} \\ & =\sqrt{(60)^2+(50-50)^2} \\ z & =60 \Omega \\ & i_0=\frac{300}{60}=5 \mathrm{~A} \\ \phi & =\tan ^{-1}\left|\frac{x_L-x_C}{R}\right|=\tan ^{-1}\left|\frac{0}{60}\right|=0^{\circ} \end{aligned} $

So $(s) \rightarrow 1$

Hence, option (A) is correct.

(P) When $K_1$ is closed current in $R_0$ is $I_1$

At $\mathrm{t}=0$; circuit will be

$\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1) \end{aligned}$

(Q) After long time inductor behave as a wire so $I_2$

$\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3) \end{aligned}$

$\text { (R) When } K_2 \text { is closed and } K_1 \text { open }$

$\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \mathrm{kilo}-\mathrm{radian} / \mathrm{s} \\ & \mathrm{R} \rightarrow(2) \end{aligned}$

(S) Now $K_2$ is closed and $K_1$ open

$\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5) \end{aligned}$

To solve this problem, we need to understand the working principle of step-up and step-down transformers and the relationships between the voltages and the number of turns of the windings. The key formula we use is the transformer equation:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where:

• $V_p$ is the primary voltage
• $V_s$ is the secondary voltage
• $N_p$ is the number of turns in the primary winding
• $N_s$ is the number of turns in the secondary winding

Let's break down the problem:

1. The power plant produces an electric power of 600 kW at 4000 V. 2. A step-up transformer is used at the plant side with a turns ratio of 1 : 10, i.e., $N_p : N_s = 1 : 10$.

This tells us that:

$\frac{V_p}{V_s} = \frac{1}{10}$

Since the primary voltage $V_p$ is 4000 V, we can find the secondary voltage $V_s$ as follows:

$V_s = 10 \times V_p$

$V_s = 10 \times 4000$

$V_s = 40000$ V

So, at the output of the step-up transformer, the voltage is 40000 V. This high voltage is then transported over the cables to the consumers' location 20 km away. At the consumers' end, a step-down transformer is used to reduce the voltage to 200 V to supply power to the consumers.

Now, we are given that the power supplied at the consumers' end must be 200 V. We need to find the ratio of the number of turns in the primary to the number of turns in the secondary for the step-down transformer so that it outputs 200 V from an input of 40000 V.

Using the transformer equation again, we can write:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where $V_p = 40000$ V and $V_s = 200$ V. Plugging these values in, we get:

$\frac{40000}{200} = \frac{N_p}{N_s}$

This simplifies to:

$\frac{40000}{200} = 200$

So, the turns ratio $\frac{N_p}{N_s} = 200 : 1$.

Thus, the correct answer is:

Option A: 200 : 1

In circuit $(P)$, under steady state, capacitor will act like an infinite impedance and inductor will act like a zero impedance. Thus, $I=0, V_1=0$, and $V_2=V$.

In circuit $(Q)$, inductor will act as zero resistance in steady state giving us $I=V / R=V / 2, V_1=0$, and $V_2=V$.

In circuit $(R)$, the inductive reactance $X_L$ and impedance $Z$ are

$ \begin{aligned} X_L & =\omega L=2 \pi \nu L=1.88 \Omega, \quad \text { and } \\\\ Z & =\sqrt{X_L^2+R^2}=2.75 \Omega \end{aligned} $

Thus, $I=V / Z \neq 0, V_1=X_L I=1.88 I, V_2=R I=2 I$, and $V_2>V_1$.

In circuit (S), inductive reactance $X_L$, capacitive reactance $X_C$, and impedance $Z$ are

$ \begin{aligned} X_L & =1.88 \Omega \\\\ X_C & =1 /(\omega C)=1061 \Omega, \quad \text { and } \\\\ Z & =X_C-X_L=1059 \Omega \end{aligned} $

Thus the current in the circuit $I=V / Z \neq 0, V_1=$ $X_L I=1.88 I, V_2=X_C I=1061 I$, and $V_2>V_1$.

In circuit $(T), X_C, R$, and $Z$ are

$ \begin{aligned} X_C & =1 /(\omega C)=1061 \Omega \\\\ R & =1000 \Omega, \quad \text { and } \\\\ Z & =\sqrt{R^2+X_C^2}=1458 \Omega \end{aligned} $

Thus, $I=V / Z \neq 0, V_1=R I=1000 I, V_2=X_C I=$ $1061 I$, and $V_2 > V_1$.

Impedance of the circuit,

$Z = \sqrt {{{({X_C})}^2} + {{(R)}^2}} = \sqrt {{{\left( {{1 \over {\omega C}}} \right)}^2} + {R^2}} $

As $\omega$ increases, Z decreases.

Current in the circuit, $I = {{{V_0}} \over Z}$

When $\omega$ is increased, the impedance of the circuit decreases and the current through the bulb increases. Therefore the bulb glows brighter.

As shown in the figure, the horizontal component of the magnetic field interacts with the current in the conducting ring, which produces an average force in the upwards direction (by using Fleming's left hand rule).

(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} $

The potential difference between point X and Y is given by

V_XY = V_X $-$ V_Y

= V₀ sin($\omega$t) $-$ V₀ sin($\omega$t + ${{2\pi } \over 3}$)

= 2V₀ cos($\omega$t + ${{\pi } \over 3}$) sin ($-$${{\pi } \over 3}$)

= $-$$\sqrt 3 $V₀ cos($\omega$t + ${\pi \over 3}$) = $\sqrt 3 $V₀ sin ($\omega$t $-$ ${\pi \over 6}$).

Similarly, the potential difference between point Y and Z is

V_YZ = V_Y $-$ V_Z

= V₀ sin($\omega$t + ${2\pi \over 3}$) $-$ V₀ sin($\omega$t + ${4\pi \over 3}$)

= 2V₀ cos($\omega$t + $\pi$) sin ($-$${\pi \over 3}$)

= $\sqrt3$V₀ cos($\omega$t) = $\sqrt3$V₀ sin($\omega$t + ${\pi \over 2}$),

and the potential difference between point Z and X is

V_ZX = V_Z $-$ V_X

= V₀ sin($\omega$t + ${4\pi \over 3}$) $-$ V₀ sin($\omega$t)

= $\sqrt3$V₀ cos($\omega$ + ${2\pi \over 3}$) = $\sqrt3$V₀ sin($\omega$t + ${7\pi \over 6}$).

The rms value of the potential V = V₀ sin($\omega$t + $\phi$) is given by V^rms = V₀ / $\sqrt2$. Thus, rms values of the potentials are $V_{XY}^{rms} = V_{YZ}^{rms} = V_{ZX}^{rms} = {V_0}\sqrt {3/2} $. Hence, the reading of the voltmeter is independent of the two terminal i.e., reading is same whether it is connected across X-Y or Y-Z or Z-X.

Given V = V₀ sin $\omega$t

At resonant frequency, current and voltage will be in same phase i.e., $\omega L = {1 \over {\omega C}}$

$\therefore$ Resonant frequency $\omega = {1 \over {\sqrt {LC} }}$ ...... (i)

Here, L = 1 $\mu$H = 10^$-$6 H, C = 10^$-$6 F

$\therefore$ $\omega = {1 \over {\sqrt {{{10}^{ - 12}}} }} = {10^6}$ rad s^$-$1

So, option (a) is incorrect.

Since, ${X_C} = {1 \over {\omega C}},\,{X_L} = \omega L$ ....... (ii)

For large $\omega$, X_C $\sim$ 0, so circuit will behave like a inductive circuit.

So, option (d) is incorrect.

From Eqn. (i)

$\omega = {1 \over {\sqrt {LC} }}$

At resonant frequency or the frequency at which the current will be in phase with the voltage is independent of R.

$\therefore$ option (b) is correct.

From Eqn. (ii), at $\omega$ $\sim$ 0, X_C $\sim$ $\infty$ and X_L = 0

$\therefore$ The current flowing through the circuit becomes nearly zero.

So, option (c) is correct.