wht's the ans..i think there is sum mistake in the values given in the Q
Q1. In an L-C circuit the values of C = 30 F and L = 15 mH. At an instant t, the charge on the capacitor is 10 C and the current is 20 mA. The maximum current in the circuit is
(A) 17 mA
(B) 19 mA
(C) 25 mA
(D) 42 mA
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3 Answers
The charge q on the capacitor is
q(t)=A\cos(\omega t+\varphi)
where \omega =\sqrt{ \dfrac{1}{LC}}.
So the current will be given by
i(t)=-A\omega \sin(\omega t+\varphi)
Let at t=t0, the charge q=q0=10 C, and i=i0=20 mA. So that we have
i_0=-A\omega \sin(\omega t_0+\varphi)
and
q_0=A\cos(\omega t_0+\varphi)
Squaring and adding, we get
A=\sqrt{q_0^2+\dfrac{i_0^2}{\omega^2}}
Accordingly the maximum current in the circuit is
i_\textrm{max}=\omega\sqrt{q_0^2+\dfrac{i_0^2}{\omega^2}}=\sqrt{\omega^2q_0^2+i_0^2}=\sqrt{\dfrac{q_0^2}{LC}+i_0^2}
Now plug in the values.