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Explain AC circuit with only capacitor. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :A pure capacitor connected in AC circuit is shown in figure. Capacitor is connected with ac voltage `V =V_(m) sin <a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a> t`. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XII_P1_C07_E01_057_S01.png" width="80%"/> <br/> When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. <br/> As charge accumulates on the capacitor <a href="https://interviewquestions.tuteehub.com/tag/plates-1156478" style="font-weight:bold;" target="_blank" title="Click to know more about PLATES">PLATES</a>, the voltage across them increases opposing the current. <br/> When the capacitor is fully charged, the current in the circuit <a href="https://interviewquestions.tuteehub.com/tag/falls-983294" style="font-weight:bold;" target="_blank" title="Click to know more about FALLS">FALLS</a> to zero. <br/> When the capacitor is connected to an across source, it limits or regulates the current but does not completely prevent the flow of charge. <br/> The capacitor is alternately charged and discharged as the current reverses each half cycle. <br/> Let q be the charge on the capacitor at any time t. The instantaneous voltage V across thecapacitor is `V = ( q)/( C )` where C is the capacitance of capacitor. <br/> From the Kirchhoff.s loop rule, <br/> `V _(m)sin omega t- ( q )/( c ) = 0` <br/> `:. V_(m) sin omegat= ( q)/( C )` <br/> `:. CV_(m) sin omega tq ` <br/> NOw `I = ( dq)/( dt )` <br/> `:. I = ( d )/( dt) [ V_(m) C sin omega t ]` <br/> `:. I = V_(m) C omega cos omega t ` <br/> `:. I = ( V _(m))/((1)/( omega C )) . cos omega t ` <br/> Now putting `cos omega t = sin ( omega t + ( pi )/( 2))` <br/>`:. I = I_(m) sin ( omega t + (pi )/( 2))` .......(2) <br/>where, `I_(m) = ( V_(m))/((1)/( omega C ))` is the amplitude of the oxcillating current.<br/> Comparing this equation to `I_(m) = ( V_(m))/( R )` <br/> `(1)/( omega C )` plays the role of resistance. It is called capacitance reactance and is denoted by `X_(C )`. <br/> `:. X_(C ) = ( 1)/( omega C )` <br/> So that the amplitude of the current is `I_(m) = ( V_(m))/( X_(C ))` <br/> SI unit of capacitance reactanceis Ohm `( Omega )` . <br/> The capacitive reactance limits the amlitude of the current in the purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. <br/> In a pure capacitive circuit amplitude of current is inversely proportional to the frequency and the capacitance. <br/> A comparison of `V = V(m) sin omega t ` and `I = I_(m) sin ( omega t + ( pi )/(2))` shows the current is `( pi )/(2)` ahead of voltage. <br/> Below figure shows the phasor diagram at an instant `t_(1)`. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XII_P1_C07_E01_057_S02.png" width="80%"/> <br/> Here the current phasor I is `( pi )/(2)` ahead of the voltage phasor V as they <a href="https://interviewquestions.tuteehub.com/tag/rotate-614912" style="font-weight:bold;" target="_blank" title="Click to know more about ROTATE">ROTATE</a> counter clockwise. ( Lead) <br/> Figure (b) shows that variation of voltage and current with time. The current reaches its maximum value earlier than the voltage by one -fourth of a period. <br/> One-fourth of a period. <br/> `(T)/(4) = ( T )/( 2pi ) xx ( 2pi )/( 4) = ( (pi )/( 2)) /( omega ) = ( pi )/( 2 omega )`</body></html> | |