Question

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.

Short answer

1. The rank of the circuit according to the current amplitude is$Circuit3>Circuit2>Circuit1$ .
2. The rank of the circuit according to the period is $Circuit2>Circuit3=Circuit1$.

Step-by-step solution

The given data

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: $q=3\cos 4t$

Understanding the concept of charge of LC circuit

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)

a) Calculation of the rank of the circuits according to the current amplitude

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

$Circuit3>Circuit2>Circuit1$

b) Calculation of the rank of the circuits according to the period

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

$Circuit2>Circuit3=Circuit1$.