Question

Charge (in $\mu$C) on capacitor as a function of time is

• A. $q=10\sqrt{2}\sin \left({10}^{4}t\right(sec{.}^{-1})+\frac{\pi }{4})$
• B. $q=10\sin \left({10}^{4}t\right(sec{.}^{-1})+\frac{\pi }{2})$
• C. $q=10\sin \left({10}^{4}t\right(sec{.}^{-1})+\frac{\pi }{4})$
• D. $q=10\sqrt{2}\sin \left(1062t\right(sec{.}^{-1})+\frac{\pi }{4})$

Answer 1

The correct option is A $q=10\sqrt{2}\sin \left({10}^{4}t\right(sec{.}^{-1})+\frac{\pi }{4})$

Since charge is in phase difference with current,
And thus the equation is in the form of $q=q_{max}\sin (\omega t+\varphi )$
Now, capacitor is fully charged initially with $10\mu C$, which is the rms value,
Thus peak value $q_{max}=10\sqrt{2}$
Also, Initial phase difference is given by,
$\tan \varphi =\frac{X_C}{X_L}=\frac{1}{{\omega }^2\times LC}=1$
$\therefore \varphi =\frac{\pi }{4}$
Also, $\omega =\sqrt{\frac{1}{LC}}={10}^4$