Alternating Current

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

Voltage across secondary coil,

= ${{Power\,across\,load} \over {Current\,pas\sin g\,through\,load}}$

$ \Rightarrow {V_2} = {P \over {{I_L}}} = {{60} \over {0.11}} \Rightarrow {V_2} = 545.45$ V

Voltage across primary coil, V₁ = 220V

$\Rightarrow$ V₂ > V₁

It means that step-up transformer is used.

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