12th Class Physics Sample Paper Physics Sample Paper-4

  • question_answer
    (i) State the condition for resonance to occur in a series L-C-R AC circuit and derive an expression for the resonant frequency. Draw the plot showing the variation of the peak current \[({{I}_{m}})\] with frequency of the AC source used. Also define the quality factor Q of the circuit.
    (ii) Calculate the (a) impedance (b) wattles current of the given AC circuit.
    (i) Mention the reasons for energy losses in an actual transformer.
    (ii) The power transmission lines needs input power at 2300 V to a step down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary windings in order to get output power at 230 V.


    (i) Resonance in Series L-C-R circuit When the frequency of the voltage source is near to the natural frequency of the L-C-R circuit, the amplitude of current is maximum. This is called resonance condition. For resonance : The voltage should be in phase with the current flowing in the circuit. Current amplitude is given by.             \[{{i}_{m}}=\frac{{{V}_{m}}}{Z}=\frac{{{V}_{m}}}{\sqrt{{{R}^{2}}+{{({{X}_{C}}-{{X}_{L}})}^{2}}}}\] where,   \[{{X}_{C}}=\frac{1}{\omega C}\] and \[{{X}_{L}}=\omega L\] If the frequency a is varied, then at a particular frequency \[{{\omega }_{0}}.\]             \[{{X}_{L}}={{X}_{C}}\Rightarrow {{\omega }_{0}}L=1/{{\omega }_{0}}C\Rightarrow \omega _{0}^{2}LC=1\] \[\therefore \] The resonant frequency, \[{{\omega }_{0}}=1\sqrt{LC}\] At \[\omega ={{\omega }_{0}},{{X}_{L}}={{X}_{C}}\Rightarrow Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=R\] \[{{i}_{m}}=\frac{{{V}_{m}}}{Z}=\frac{{{V}_{m}}}{R}\] So, at resonance frequency, the impedance is minimum and current amplitude is maximum. Plot showing the variation of peak current \[({{i}_{m}})\] with the frequency of the ac source \[(\omega ):\] Quality factor (Q) The quality factor of Q is defined as             \[Q=\frac{{{\omega }_{0}}L}{R}=\frac{1}{{{\omega }_{0}}CR}\] The quality factor is an indicator of the sharpness of the resonance. The higher value of Q indicates sharper peak in the current. (ii) In the given circuit, I = 2 A             \[{{V}_{C}}=40\,V=I{{X}_{C}}\] or \[{{C}_{R}}=30\,V=IR\] \[\Rightarrow \]   \[40=2\times {{X}_{c}}\Rightarrow {{X}_{c}}=20\,\Omega \] \[\Rightarrow \]   \[30=2\times R\Rightarrow R=15\,\Omega \] (a) Impedance, \[Z=\sqrt{{{R}^{2}}+{{X}_{c}}^{2}}=\sqrt{{{15}^{2}}+{{20}^{2}}}=25\,\Omega \] (b) Wattless current \[=I\sin \phi \] \[\phi ={{\tan }^{-1}}\left( \frac{{{X}_{c}}}{R} \right){{\tan }^{-1}}\left( \frac{20}{15} \right)=53.13\] \[\therefore \]   Wattless current \[=2\times \sin (53.13)=1.6A\] Or

    (i) Reasons for energy losses in actual transformer
    (a) Joule heating Energy is lost in resistance of primary and secondary winding as heat \[({{I}^{2}}Rt)\] is generated.
    (b) Flux leakage Energy is lost due to coupling of primary and secondary coils not being perfect, i.e. whole of magnetic flux generated in primary coil is not linked with the secondary coil.
    (c) Eddy currents The alternating magnetic flux induces eddy currents in the iron core and causes heating.
    (d) Hysteresis loss This is the loss of energy due to repeated magnetisation and demagnetization because of AC in it.
    (ii) Given, \[{{V}_{P}}=2300\,V,\] \[{{N}_{P}}=4000\,turns,\] \[{{V}_{S}}=230\,V\] and \[{{N}_{S}}=?\]
    We know that, \[{{V}_{S}}/{{V}_{P}}={{N}_{S}}/{{N}_{P}}\] \[\Rightarrow \]   \[{{N}_{S}}=\frac{{{V}_{S}}}{{{V}_{P}}}\times {{N}_{P}}=\frac{230}{2300}\times 4000\]             = 400 turns

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