Alternating Current

$ \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

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

$\begin{aligned} & \omega_0=10^5 \mathrm{rad} \mathrm{s}^{-1}, L=50 \times 10^{-3} \mathrm{H} \\\\ & \Rightarrow X_L=5000 \Omega=X_C \text { (at resonance) } \\\\ & \frac{1}{\sqrt{L C}}=\omega_0 \\\\ & \Rightarrow L_C=\frac{1}{\omega_0^2} \\\\ & \Rightarrow C=2 \times 10^{-9} \mathrm{~F} \\\\ & I_0=\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\\\ & I_0=\frac{\varepsilon_0}{R} ..........(i) \end{aligned}$

$ \begin{aligned} 0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} ..........(ii) \\\\ \text { At } \omega & =8 \times 10^4 \mathrm{rad} \mathrm{s}^{-1} \\\\ X_L & =8 \times 10^4 \times 50 \times 10^{-3}=4000 \Omega \\\\ X_C & =\frac{1}{\omega_C}=6250 \Omega \end{aligned} $

From (i) and (ii), we get

$ \begin{aligned} & \sqrt{R^2+\left(X_L-X_C\right)^2} =\frac{R}{0.05} \\\\ & R^2+(6250-4000)^2 =400 R^2 \\\\ & \Rightarrow R^2 \simeq \frac{(2250)^2}{400} \\\\ & R =\frac{2250}{20}=112.5 \Omega \\\\ &I_0 =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\\\ & =400 \mathrm{~m} \mathrm{~A} \end{aligned} $

$(\mathrm{P} \rightarrow 3)$

$\begin{aligned} \rightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\\\ \Rightarrow Q & =\frac{2}{2250} \times \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-7}}}=\frac{10^4}{225} \\\\ & =44.4\\\\ (\mathrm{Q} \rightarrow 1) \\\\ Q & =\frac{\omega_0}{B W}\end{aligned}$

$\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250$

$(\mathrm{R} \rightarrow 4)$

$\begin{array}{rrr}P =\frac{\varepsilon_0^2}{R} \text { (peak power) } \\\\ =\frac{45^2}{112.5}=18 \mathrm{~W} \end{array}$

$(S \rightarrow 2)$

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.

The current flowing through the capacitor and its charge are given by

$i = {i_0}\cos \omega t$, .... (1)

$Q = \int {idt = \int {{i_0}\cos \omega t\,dt = {{{i_0}} \over \omega }\sin \omega t + k} } $ $ = {{{i_0}} \over \omega }\sin \omega t$, ...... (2)

where we have used the initial condition, Q = 0 at t = 0, to get the integration constant k = 0. The charge attains the maximum value Q_max at t = $\pi$ / (2$\omega$),

Q_max = i₀/$\omega$ = 1/500 = 2 $\times$ 10^$-$3 C.

The key is switched from B to D at the time t = 7$\pi$/6$\omega$. Substitute t = 7$\pi$/6$\omega$ in equations (1) and (2) to get current and charge

i = i₀ cos(7$\pi$/6) = cos($\pi$ + $\pi$/6) = $-$ cos($\pi$/6)

$ = - \sqrt 3 /2$ A,

Q_i = (i₀/$\omega$) sin(7$\pi$/6) = 2 $\times$ 10^$-$3 sin($\pi$ + $\pi$/6)

= $-$ 1 $\times$ 10^$-$3 C.

Note that negative sign in i indicates the counterclockwise direction of current and negative sign in Q_i indicates the negative charge on the upper plate.

The potential across the capacitor is ${V_C} = {{{Q_i}} \over C} = - {{1 \times {{10}^{ - 3}}} \over {20 \times {{10}^{ - 6}}}}$ = $-$ 50 V (upper plate at lower potential). Immediately after A is connected to D, the potential across the resistor is V_R = 100 V and hence the current through it is i = V_R/R = 100/10 = 10 A (counterclockwise). The potential across the capacitor when it is fully charged is equal to the battery emf. Thus, the charge on the capacitor in fully charged condition is

Q_f = V/C = 50/(20 $\times$ 10^$-$6) = 1 $\times$ 10^$-$3 C.

The sign of Q_f indicates the positive charge on the upper plate. Hence, charge that flows through the battery is

Q = Q_f $-$ Q_i = 1 $\times$ 10^$-$3 $-$ ($-$ 1 $\times$ 10^$-$3) = 2 $\times$ 10^$-$3 C.

Here, $\Omega$ = 100 rad/s, L = 0.5 H, C = 100 $\mu$F, V = 20 V

$\therefore$ ${X_L} = \omega L = 100 \times 0.5 = 50\,\Omega $

${X_C} = {1 \over {\omega C}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 100\,\Omega $

Impedance across capacitor,

${Z_1} = \sqrt {{R^2} + X_C^2} $

$ = \sqrt {{{(100)}^2} + {{(100)}^2}} $

${Z_1} = 100\sqrt 2 \,\Omega $

$\therefore$ ${I_1} = {{20} \over {100\sqrt 2 }} = {1 \over {5\sqrt 2 }}\,A$

Voltage across 100 $\Omega$

$V = {I_1} \times 100 = {1 \over {5\sqrt 2 }} \times 100 = 10\sqrt 2 \,V$

Impedance across inductance,

${Z_2} = \sqrt {{R^2} + {{({X_L})}^2}} = \sqrt {{{(50)}^2} + {{(50)}^2}} $

${Z_2} = 50\sqrt 2 \,\Omega $

$\therefore$ ${I_2} = {{20} \over {50\sqrt 2 }} = {2 \over {5\sqrt 2 }} = {{\sqrt 2 } \over 5}$

Now, voltage across $50\,\Omega = {{\sqrt 2 } \over 5} \times 50 = 10\sqrt 2 $

${I_1} = {1 \over {5\sqrt 2 }}A$ at 45$^\circ$ leading

${I_2} = {{\sqrt 2 } \over 5}A$ at 45$^\circ$ lagging

$\therefore$ Current through circuit

${I_{net}} = \sqrt {I_1^2 + I_2^2} = \sqrt {{{\left( {{1 \over {5\sqrt 2 }}} \right)}^2} + \left( {{{\sqrt 2 } \over 5}} \right)} = 0.3\,A$

In case A,

The capacitive reactance is

$X_C^A = {1 \over {\omega C}}$

Impedance of the circuit is

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

$I_R^A = {V \over {\sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} }}$ ...... (i)

$V_C^A = {{I_R^A} \over {\omega C}} = {V \over {\sqrt {{{(R\omega C)}^2} + 1} }}$ ...... (ii)

In case B,

The capacitive reactance is

$X_C^B = {1 \over {\omega (4C)}} = {1 \over {4\omega C}}$

Impedance of the circuit is

${Z_B} = \sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} $

$I_R^B = {V \over {\sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} }}$ ..... (iii)

$V_C^B = {V \over {\sqrt {{{(4R\omega C)}^2} + 1} }}$ ..... (iv)

From (i) and (iii), we conclude that

$I_R^A < I_R^B$

From (ii) and (iv), we conclude that

$V_C^A > V_C^B$