Figure 31-20 shows graphs of capacitor voltage ${\mathit{V}}_{\mathbf{C}}$for $\mathit{L}\mathit{C}$
circuits 1 and 2, which contain identical capacitances and have the same, maximum charge Q. Are (a) the inductance L and (b) the maximum current I in circuit 1 greater than, less than, or the same as those in circuit?

1. The capacitors in the two circuits are identical.
2. The two capacitors have the same maximum charge Q.

The period of oscillations in the LC circuit depends upon the value of the inductance and capacitance in the circuit. The maximum current in the circuit also depends upon these quantities. Thus, we can rank the two circuits by comparing the value of inductances as the capacitors are identical.

Formulae:

The angular frequency of an oscillation,$\mathit{\omega }\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{T}}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{L}\mathbf{C}}}\mathbf{}$ (i)

The current in the LC circuit, $\mathit{I}\mathbf{=}\mathit{\omega }\mathbf{}\mathit{Q}\mathbf{\text{or}}\mathit{I}\mathbf{=}\mathbf{}\frac{\mathbf{Q}}{\sqrt{\mathbf{L}\mathbf{C}}}$ (ii)

Observing the graph, we can see that the frequency of the circuit 1 is more than that of circuit 2. So the period of oscillation is less for circuit 1 than that circuit 2.

The period of oscillation relation to the capacitance can be given using equation (i) as follows:

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ =2\pi \sqrt{LC} \end{gathered}$$

Thus, we see that $T\propto \sqrt{LC}.$

Since the capacitance in the two circuits is identical, the period dependence can be given as:$T\propto \sqrt{L}.$

Now, the period isless for circuit 1 than that circuit 2.

Hence the inductance in circuit 1 must be less than that in circuit 2.

The capacitance and the maximum charge Q are identical in both the circuits, so the maximum current will depend on the inductance value only (from equation (ii)), i.e.,

.$I\propto \frac{1}{\sqrt{L}}$

For circuit 1, the inductance is lesser than that in circuit 2.

Hence the maximum current in circuit 1 will be greater than that in circuit 2.