Charges on the capacitors in three oscillating LC circuits vary as: (1) $\mathit{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathit{t}$ , (2)$\mathit{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{t}$ , (3)$\mathit{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathit{t}$ (with q in coulombs and t in seconds). Rank the circuits according to (a) the current amplitude and (b) the period, greatest first.

The charges of the capacitors in three oscillating LC circuits vary as:

Circuit 1: $q=2\cos 4t$

Circuit 2: $q=4\cos t$

Circuit 3: $q=3\cos 4t$

The LC circuit has an oscillating charge on the capacitor. This indicates that the current through the circuit also varies. The oscillations of the charge and current depend on the values of the period of oscillation.

Formulae:

The charge of the capacitor varies as $\mathit{q}\mathbf{=}\mathit{Q}\mathbf{}\mathbf{\text{cos}}\left(\omega t+\varphi \right)$ (i)

The current amplitude of the LC circuit, $\mathit{I}\mathbf{=}\mathit{\omega }\mathit{Q}$ (ii)

The period of an oscillation, $\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\omega }}$ (iii)

So for $\varphi =0$, we can say using equation (i) the charge will vary as $q=Q\text{cos}\left(\omega t\right)$

For circuit 1, $Q=2$and $\omega =4$

Thus, the value of the current amplitude using the data in equation (ii) can be given as:

$I=\omega Q$

For circuit 2, Q = 4 and$\omega =1$

Thus, the value of the current amplitude using the data in equation (ii) can be given as:

$I=4\text{A}$

For circuit 3, Q = 3 and $\omega =4$

Thus, the value of the current amplitude using the data in equation (ii) can be given as:

$I=12\text{A}$

Hence, the rank of the circuits according to current amplitude is

$Circuit3>Circuit2>Circuit1$

For circuit 1, Q = 2 and $\omega =4$

Thus, the period of the oscillation using the data in equation (iii) can be given as follows:
$$\begin{gathered} T=\frac{2\pi }{4} \\ =\frac{\pi }{2} \end{gathered}$$

For circuit 2, Q = 4 and$\omega =1$

Thus, the period of the oscillation using the data in equation (iii) can be given as follows:
$$\begin{gathered} T=\frac{2\pi }{1} \\ =2\pi \end{gathered}$$

For circuit 3, Q = 3 and$\omega =4$

Thus, the period of the oscillation using the data in equation (iii) can be given as follows:
$$\begin{gathered} T=\frac{2\pi }{4} \\ =\frac{\pi }{2} \end{gathered}$$

Hence, the rank of the circuits according to their period of oscillation is

$Circuit2>Circuit3=Circuit1$.