question_answer

An emf \[E=100\,sin\,314\,t\] is applied across a pure capacitor of \[637\mu F.\] Find

(i) the instantaneous current I

(ii) instantaneous power P

(iii) the frequency of power and

(iv) the maximum energy stored in the capacitor.

Answer:

(i) Given, \[E=100\,sin\,314\,t\,volt\] As the current in a capacitor leads the voltage by \[90{}^\circ ,\] so the instantaneous current is given by             \[I={{I}_{0}}\sin (314\,t+90{}^\circ )={{I}_{0}}\cos 31\,4t\] where,   \[{{I}_{0}}=\frac{{{E}_{0}}}{{{X}_{C}}}=\frac{{{E}_{0}}}{1/\varepsilon C}={{E}_{0}}\omega C\] But, \[{{E}_{0}}=100V,\] \[\omega =314\,rad\,{{s}^{-1}},\] \[C=637\times {{10}^{-6}}F\] \[\therefore \]      \[{{I}_{0}}=100\times 314\times 637\times {{10}^{-6}}=20A\] Hence,  \[I=20cos\text{ }314\,t\] ampere. (ii) Instantaneous power,             \[P=EI=100\sin 314\,t\times 20\cos 314\,t\]             \[=1000\sin 628\,t\,watt\] (iii) Angular frequency of power, \[{{\omega }_{p}}=628\,rad\,{{s}^{-1}}\] \[\therefore \] Frequency of power, \[{{f}_{H}}=\frac{{{\omega }_{H}}}{2\pi }=\frac{628}{2\pi }=100Hz\] (iv) The maximum energy stored in the capacitor is \[{{U}_{0}}=\frac{1}{2}CE_{0}^{2}=\frac{1}{2}\times 637\times {{10}^{-6}}\times {{(100)}^{2}}=3.185\,J\]