Alternating Current

As V_L = V_C = 300 V, resonance will take place

$ \therefore $ V_R = 220 V

Current, I = ${{220} \over {100}}$ = 2.2 A

$ \therefore $ Hence, the reading of the voltmeter V₃ is 220 V and the reading of ammeter A is 2.2 A.

Power dissipated, P = E_rmsI_rms cos$\phi $

cos$\phi $ = ${R \over Z}$

But I_rms = ${{{E_{rms}}} \over Z}$

$ \therefore $ P = $E_{rms}^2.{R \over {{Z^2}}}$

Also we know, Z = $\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $

So, P = $E_{rms}^2.{R \over {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}}$

= ${{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}}$

Average power = ${{{E_0}{I_0}} \over 2}\cos \phi $

Efficiency of the transformer

= ${{{P_{output}}} \over {{P_{input}}}} \times 100$

= ${{100} \over {220 \times 0.5}} \times 100$ = 90 %

Voltage across the primary,

V_p = ${{d\phi } \over {dt}} = {d \over {dt}}\left( {{\phi _0} + 4t} \right) = $ 4 volt

Also ${{{V_s}} \over {{V_p}}} = {{{N_s}} \over {{N_p}}}$

Where, Ns = No. of turns across primary coil = 50

N_p = No. of turns across secondary coil = 1500

$ \therefore $ V_s = ${{1500} \over {50}} \times 4$ = 120 V

Condition for which current is maximum in a series LCR circuit is,

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

$ \Rightarrow $ 1000 = ${1 \over {\sqrt {L\left( {10 \times {{10}^6}} \right)} }}$

$ \Rightarrow $ L = 100 mH

Impedance of series LCR is

Z = $\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $

= $\sqrt {{8^2} + {{\left( {31 - 25} \right)}^2}} $

= $\sqrt {64 + 36} $ = 10 $\Omega $

Power factor, cos$\phi $ = ${R \over Z} = {8 \over {10}} = 0.8$

We know that frequency of electrical oscillation in L.C. circuit is

f = ${1 \over {2\pi }}\sqrt {{1 \over {LC}}} $

$ \therefore $ ${{{f_1}} \over {{f_2}}} = {\left( {{{{L_2}{C_2}} \over {{L_1}{C_1}}}} \right)^{1/2}}$

= ${\left( {{{2L \times 4C} \over {L \times C}}} \right)^{1/2}} = {\left( 8 \right)^{1/2}}$

$ \therefore $ ${{{f_1}} \over {{f_2}}} = 2\sqrt 2 $

$ \Rightarrow $ ${{f_2} = {{{f_1}} \over {2\sqrt 2 }}}$

$ \Rightarrow $ ${{f_2} = {f \over {2\sqrt 2 }}}$ [As f₁ = f]

The core of a transformer is laminated to minimise the energy losses due to eddy currents.

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

$ \Rightarrow $ $\tan \left( {{\pi \over 4}} \right) = {{{1 \over {\omega C}} - \omega L} \over R}$

$ \Rightarrow $ R = ${{1 \over {\omega C}} - \omega L}$

$ \Rightarrow $ R + 2$\pi $fL = ${1 \over {2\pi fC}}$

$ \Rightarrow $ C = ${1 \over {2\pi f\left( {R + 2\pi fL} \right)}}$

Time constant of LR circuit is $\tau $ = L/R.

$ \therefore $ $\tau $ = 40/8 = 5 sec.

The impedance Z of a series LCR circuit is given by,

Z = $\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $

At resonance, X_L = X_C, hence Z = R.

Let, supply voltage = V_R = V

$ \therefore $ R.M.S. current, I = ${V \over R}$

$ \therefore $ Power loss = VI = I²R

Reactance X = ${1 \over {\omega C}} = {1 \over {2\pi fC}}$

$ \therefore $ X $ \propto $ ${1 \over {fC}}$

$ \therefore $ ${{{X_1}} \over X} = {{fC} \over {{f_1}{C_1}}}$ = ${{fC} \over {2f \times 2C}}$ = ${1 \over 4}$

$ \Rightarrow $ X₁ = ${X \over 4}$

Quality factor Q = ${{\omega L} \over R}$

As ${\omega ^2} = {1 \over {LC}}$

$ \therefore $ Q = ${1 \over {\omega RC}}$