Alternating Current

Given:

Peak current = 2 A

Peak voltage = 1 V

Power factor = $ \frac{1}{2} $

The power factor is defined as the ratio of real power to apparent power. For an AC circuit, it can also be described as $ \cos \phi $, where $ \phi $ is the phase angle between the current and the voltage.

Here is the derivation:

$ \text{Power factor} = \frac{\text{Peak voltage}}{\text{Peak current}} = \frac{1}{2} $

Since the power factor is also expressed as:

$ \text{Power factor} = \cos \phi $

We set up the equation:

$ \cos \phi = \frac{1}{2} $

Solving for $ \phi $, we get:

$ \phi = \cos^{-1}\left(\frac{1}{2}\right) $

This leads to:

$ \phi = 60^\circ $

Therefore, the angle between the voltage and current is $ 60^\circ $.

$I = {V \over {\omega L}}$

$ = {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}$

$ \Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A$

$i(t) = {V \over R}(1 - {e^{ - Rt/L}})$ ...... (1)

${L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms$ ...... (2)

${V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})$

$ \Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms$

$U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} \right]^2}$

$ = {1 \over 2}{[1 - 0.25]^2} \times 36 \times {10^{ - 4}}$

$ = 1\,mJ$

${V_1}{i_1} = {V_2}{i_2} = 80$ kW

$ \Rightarrow {i_1} = 10\,A$ and ${i_2} = {{80 \times 1000} \over {160}} = 500\,A$

$ \Rightarrow {R_1} = {{{V_1}} \over {{i_1}}} = 800\,\Omega $ and ${R_2} = {{160} \over {500}} = 0.32\,\Omega $

$V(t) = I\omega L\sin (49\pi t - 30^\circ + 90^\circ )$

$ = 5 \times 49\pi \times {{30} \over {1000}}\sin (49\pi t + 60^\circ )$

$ = 23.1\sin (49\pi t + 60^\circ )$

$\omega = 0.4{\omega _0}$ ...... (i)

$ \Rightarrow I = {V \over Z} = {{50} \over {\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} }}$ ..... (ii)

$ \Rightarrow I = 238$ mA

Ratio = ${{i_1^2{R_1}} \over {{{\left( {{{{i_2}} \over {\sqrt 2 }}} \right)}^2}{R_2}}} = {{{4^2} \times 3} \over {{{\left( {{4 \over {\sqrt 2 }}} \right)}^2} \times 2}}$

$\Rightarrow$ Ratio = 3 : 1

${P_1} = \cos \phi = {1 \over {\sqrt 2 }}({X_L} = R)$

${P_2} = \cos \phi ' = 1$ (will become resonance circuit)

So, ${{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$

Metal detector works on the principle of resonance in ac circuits.

Resonant frequency $ = {1 \over {\sqrt {LC} }} = {\omega _0}$

$\Rightarrow$ If we decrease C, $\omega$₀ would increase

$\Rightarrow$ Another capacitor should be added in series.

We know,

Quality factor (Q factor)

${Q_1} = {{{w_1}} \over {\Delta w}}$

$ = {1 \over {\sqrt {LC} }} \times {L \over R}$

$ = {1 \over R}\sqrt {{L \over C}} $

Now, when $L' = 2L$ and $C' = 2C$ then ${Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}$

$\therefore$ Q₂ remains same as Q₁.

Also, as ${w_1} = {1 \over {\sqrt {LC} }}$

$ \Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}$

$ \Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}$

$\therefore$ When $L' = 2L$ and $C' = 2C$ then new resonating frequency

${f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}$

We know that ${X_C} = {1 \over {\omega C}}$ and ${X_L} = \omega L$

Also, at $\omega = {\omega _r}:{X_L} = {X_C}$

$\Rightarrow$ For $\omega < {\omega _r}$ : capacitive

and $\omega = {\omega _r}:z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = R$

$U = {1 \over 2}L{i^2} = {1 \over 2}C{V^2}$

So, $\left[ {{L \over C}} \right] = {{{V^2}} \over {{i^2}}} = {R^2}$ is not the dimension of time.

$\omega = 120\pi $

$ \Rightarrow T = {1 \over {60}}\sec $

The current will take its peak value in ${T \over 4}$ time

So $t = {T \over 4}$

$ = {1 \over {240}}s$

${X_L} = 3000 \times 10 \times {10^{ - 3}} = 30\,\Omega $

${X_C} = {1 \over {3000 \times 25}} \times {10^6} = {{40} \over 3}\,\Omega $

So ${X_L} - {X_C} = 30 - {{40} \over 3} = {{50} \over 3}\,\Omega $

$\tan \theta = {{{X_L} - {X_C}} \over R} = {{50/3} \over {100}} = {1 \over 6}$

So $\theta = {\tan ^{ - 1}}(0.17)$

For wattless current to flow in AC circuit the circuit will be Purely Inductive circuit.

$X = |{X_C} - {X_L}|$

So, it can be zero if ${X_C} = {X_L}$

And, average power in ac circuit can be zero.

$I = {I_0}\cos (\omega t)$ say

$\Rightarrow$ At maximum $\omega {t_1} = 0$ or ${t_1} = 0$

Then at rms value $I = {I_0}/\sqrt 2 $

$ \Rightarrow \omega {t_2} = \pi /4$

$ \Rightarrow \omega ({t_2} - {t_1}) = \pi /4$

$\Delta t = {\pi \over {4\omega }} = {{\pi T} \over {4 \times 2\pi }}$

$ = {1 \over {400}}s$ or 2.5 ms

Given, $V_1=4000 \mathrm{~V}, I_1=100 \mathrm{~A}$,

$ V_2=2 \times 10^5 V, I_2=? $

Since, $\frac{V_1}{V_2}=\frac{I_2}{I_1}$

$ \Rightarrow \quad I_2=\frac{V_1}{V_2} \times I_1=\frac{4000}{2 \times 10^5} \times 100=2 \mathrm{~A} $

Input power to the transmission line,

$ P_i=V_2 I_2 $

$ =2 \times 10^5 \times 2=4 \times 10^5 \mathrm{~W} $

Power loss in transmission line,

$ \begin{aligned} P^{\prime} & =I_2^2 R \\ & =2^2 \times 50=200 \mathrm{~W} \end{aligned} $

Percentage of power loss in transmission line,

$ \begin{aligned} & =\frac{P^{\prime}}{P_i} \times 100 \% \\ & =\frac{200}{4 \times 10^5} \times 100 \%=\frac{1}{20} \% \\ & =0.05 \% \end{aligned} $

Capacitance of capacitor,

$ C=100 \mu \mathrm{~F}=10^{-4} \mathrm{~F} $

Resistance of coil, $R=20 \pi$

Inductance, $L=12.5 \mathrm{mH}=1.25 \times 10^{-4} \mathrm{H}$

Supply voltage, $V=220 \mathrm{~V}$

Frequency of AC source,

$ f=\frac{200}{\pi} \mathrm{~Hz} $

For maximum value of alternating current in $L-C-R$ circuit occurs at resonant condition.

i.e $X_L=X_C$

$ \begin{aligned} \therefore \quad Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ & =\sqrt{R^2+0^2} \quad\left(\because X_L=X_C\right) \\ & =R=20 \Omega \\ \therefore \quad I_{\max } & =\frac{V}{Z}=\frac{220}{20}=11 \mathrm{~A} \end{aligned} $

$ \begin{aligned}Given,\,\, L & =2 \mathrm{H}, C=32 \mu \mathrm{~F} \\ & =32 \times 10^{-6} \mathrm{~F} \\ R & =20 \Omega \end{aligned} $

$\therefore Q$-value of $L-C-R$ series circuit is given as

$ \begin{aligned} Q & =\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{20} \sqrt{\frac{2}{32 \times 10^{-6}}} \\ & =\frac{1}{20} \sqrt{\frac{1}{16 \times 10^{-6}}} \\ & =\frac{1}{20} \times \frac{1}{4 \times 10^{-3}}=\frac{10^3}{80}=\frac{100}{8} \\ & =\frac{50}{4}=12.5 \end{aligned} $

Given, $R=100 \Omega, \varepsilon=10 \sin (250 \pi) t \therefore c_0=10 \mathrm{~V}$

Angular frequency, $\omega=250 \pi \mathrm{rad} / \mathrm{s}$ Energy dissipated as heat,

$ H=\frac{\varepsilon_{\mathrm{ms}}^2}{R} \cdot t $

Energy dissipated as heat during $t=0 $

$ \begin{aligned} H & =\int_0^{10^{-3}} \frac{\varepsilon_{\mathrm{ms}}^2}{R} d t \\ & =\int_0^{10^{-3}} \frac{\varepsilon_0^2 \sin ^2 \omega t}{R} d t \\ & =\int_0^{10^{-3}} \frac{10^2}{100}\left(\frac{1-\cos 2 \omega t}{2}\right) d t \\ & =\int_0^{10^{-3}}\left[\frac{1}{2}-\frac{\cos 2 \omega t}{2}\right] d t \\ & =\left[\frac{1}{2} t\right]_0^{10^{-3}}-\left[\frac{\sin 2 \omega t}{2 \times 2 \omega}\right]_0^{10^{-3}} \\ & =\frac{1}{2} \times 10^{-3}-\frac{1}{4}\left[\sin \left(2 \times 250 \pi \times 10^{-3}\right)\right] \end{aligned} $

$ \begin{aligned} & =\frac{10^{-3}}{2}-\frac{1}{4 \omega}\left[\sin \frac{\pi}{2}\right] \\ & =\frac{10^{-3}}{2}-\frac{1}{4 \omega} \\ & =\frac{10^{-3}}{2}-\frac{1}{4 \times 250 \pi} \\ & =\frac{10^{-3}}{2}-\frac{1}{1000 \pi} \\ & =\frac{10^{-3}}{2}-\frac{10^{-3}}{\pi} \\ & =10^{-3}\left(\frac{1}{2}-\frac{1}{\pi}\right)=10^{-3}\left(\frac{\pi-2}{2 \pi}\right) \\ & =10^{-3}\left(\frac{314-2}{2 \pi}\right) \\ & =10^{-3} \times \frac{0.57}{\pi} \mathrm{~J} \end{aligned} $

$ =\frac{0.57}{\pi} \mathrm{~mJ} $

Capacitance of capacitor,

$ \begin{aligned} & C=2 \mu \mathrm{~F}=2 \times 10^{-6} \mathrm{~F} \\ & V=50 \mathrm{~V} \\ & L=10 \mathrm{mH}=10^{-2} \mathrm{H} \end{aligned} $

Potential difference across the fully charged capacitor, $V^{\prime}=V=50 \mathrm{~V}$.

∴ Induced emf, $e=V^{\prime}=V=50 \mathrm{~V}$

We know that, induced emf across the coil, $e=L \frac{d I}{d t} \Rightarrow \frac{d I}{d t}=\frac{e}{L}$

$ \begin{aligned} \Rightarrow \quad \frac{d I}{d t} & =\frac{V}{L} \quad[\because e=V] \\ & =\frac{50}{10^{-2}}=5000 \mathrm{~A} / \mathrm{s} \end{aligned} $

Expression of AC current is given by $I(t)=50 \sin (200 \pi t) \mathrm{A}$

∴ Maximum value of alternating current,

$ I_0=50 \mathrm{~A} $

∴ RMS value of alternating current,

$ I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} \mathrm{~A} $

From the given expression of alternating current, we get

$ \begin{gathered} \omega=200 \pi \\ \Rightarrow \quad 2 \pi f=200 \pi \Rightarrow \quad f=100 \mathrm{~Hz} \end{gathered} $

Hence, $I_{\text {rms }}=25 \sqrt{2}$ and $f=100 \mathrm{~Hz}$

To determine the phase angle $\phi$ between the current and the applied voltage in an $R-L-C$ circuit, we need to use the following relationship:

$\tan \phi = \frac{X_L - X_C}{R}$

where:

• $X_L$ is the inductive reactance


• $X_C$ is the capacitive reactance


• $R$ is the resistance

Given:

• Resistance, $R = 150 \Omega$
• Capacitance, $C = 20 \mu \mathrm{F} = 20 \times 10^{-6} \mathrm{F}$
• Inductance, $L = 500 \mathrm{mH} = 500 \times 10^{-3} \mathrm{H}$
• Angular frequency, $\omega = 400 \mathrm{rad} \, \mathrm{s}^{-1}$

The inductive reactance $X_L$ is calculated as:

$X_L = \omega L$

Substituting the values:

$X_L = 400 \times 500 \times 10^{-3}$

$X_L = 200 \, \Omega$

The capacitive reactance $X_C$ is calculated as:

$X_C = \frac{1}{\omega C}$

Substituting the values:

$X_C = \frac{1}{400 \times 20 \times 10^{-6}}$

$X_C = 125 \, \Omega$

Now, we can find the phase angle using the equation:

$\tan \phi = \frac{X_L - X_C}{R}$

Substituting the values:

$\tan \phi = \frac{200 - 125}{150}$

$\tan \phi = \frac{75}{150}$

$\tan \phi = 0.5$

Therefore, the phase angle $\phi$ is:

$\phi = \tan^{-1}(0.5)$

So, the correct answer is:

Option D $\tan^{-1}(0.5)$

Given capacitive reactance,

$X_{C_1}=3 \mathrm{k} \Omega$

we know that,

$X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$

$ \begin{aligned} &\begin{aligned} \Rightarrow X_C & \propto \frac{1}{f} \Rightarrow \frac{X_{C_2}}{X_{C_1}}=\frac{f_1}{f_2} \\ & =\frac{f_1}{2 f_1}=\frac{1}{2} \quad \left(\because f_2=2 f_1\right)\\ \Rightarrow X_{C_2} & =\frac{X_{C_1}}{2}\\ & =\frac{3 \mathrm{k} \Omega}{2}=1.5 \mathrm{k} \Omega \end{aligned}\\ \end{aligned}$

Given, Inductance, $L=0.1 \mathrm{~H}$

Resistance, $R=110 \Omega$;

Potential difference, $V=110 \mathrm{~V}$

Frequency, $f=350 \mathrm{~Hz}$

Phase difference between maximum voltage and maximum current is given as

$ \begin{aligned} \tan \phi & =\frac{X_L}{R} \\ & =\frac{2 \pi f L}{R}=\frac{2 \times \frac{22}{7} \times 350 \times 01}{110}=2 \\ \Rightarrow \quad \tan \phi & =2 \\ \Rightarrow \quad \phi & =\tan ^{-1}(2) \end{aligned}$