Alternating Current

The quality factor of a $R-L-C$ circuit is

$ Q=\frac{1}{R} \sqrt{\frac{L}{C}} $

At resonance, $X_L=X_C$

$ \begin{aligned} \Rightarrow & & \omega_0 L & =\frac{1}{\omega_0 C} \\ \Rightarrow & & \omega_0 & =\frac{1}{\sqrt{L C}} \\ & \text { or } & \sqrt{L / C} & =\omega_0 L \\ & \therefore & Q & =\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{\omega_0 L}{R} \end{aligned} $

Impedance ( $Z$ ) of a series $L-C-R$ circuit is given as

$ \begin{align*} Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & =\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \\ \Rightarrow \quad Z & =\sqrt{R^2+\left(2 \pi f L-\frac{1}{2 \pi f C}\right)^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

At $X_L=X_C$, then from Eq. (i), $Z_{\text {min }}=R$, and $f=f_0 \quad\left[f_0=\right.$ resonance frequency] Hence, the variation of $Z$ with frequency $f$ in a series $L-C-R$ circuit is given as

Given, applied emf is

$ E=30 \sin 200 t $

In given circuit,

$ L=0.05 \mathrm{H}, C=500 \mu \mathrm{~F} $

$ R=10 \Omega, \omega=200 \mathrm{rad} \mathrm{~s}^{-1} $

Now, inductive reactance,

$ X_L=L \omega=0.05 \times 200=10 \Omega $

Also, capacitive reactance,

$ X_C=\frac{1}{C \omega}=\frac{1}{500 \times 10^{-6} \times 200}=10 \Omega $

As $X_L=X_C$, circuit is in resonance, so amplitude of current is

$ I_{\max }=\frac{E_{\max }}{R}=\frac{30}{10}=3 \mathrm{~A} $

As coil 2 with inductance L₂ has been wounded by a similar wire as coil 1 but the direction of windings is reversed in each layer, so flux through L₂ will be zero.

As, L₂ $\propto$ $\phi$ $\Rightarrow$ L₂ = 0

Also, since the dimensions, number of turns and number of layers of windings are equal in coil 1 and 3.

So, L₁ = L₃

Given, $\omega$ = 2000 rad s^$-$1

L = 5 mH = 5 $\times$ 10^$-$3 H

C = 50 $\mu$F = 50 $\times$ 10^$-$6 F

V = 30 V

Inductive reactance,

X_L = $\omega$L = 2000 $\times$ 5 $\times$ 10^$-$3 = 10 $\Omega$

and capacitive reactance,

${X_C} = {1 \over {\omega C}} = {1 \over {50 \times {{10}^{ - 6}} \times 2000}} = {{100} \over {10}} = 10\Omega $

$\Rightarrow$ X_L = X_C

The given circuit is resonant circuit.

Hence, impedance of circuit, Z = R = 4 + 6 + 0.5 = 10.5$\Omega$

Amplitude of current, ${I_0} = {V \over Z} = {{30} \over {10.5}} = 2.85$ A

In an AC resistive circuit, current and voltage are in phase.

So, $I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)$ ..... (i)

$\therefore$ Time period of one complete cycle of current is

$T = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s$

So, current reaches its maximum value at

${t_1} = {T \over 4} = {1 \over {200}}s$

When current is half of its maximum value, then from Eq.(i), we have

$I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})$

$ \Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}$

So, instantaneous time at which current is half of maximum value is ${t_2} = {1 \over {120}}s$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3$ ms

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 $

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.

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$

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

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

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

$V_C^A > V_C^B$

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).