91影视

1. The inductors and capacitors in the three circuits are identical.
2. The two capacitors in the circuit b are in parallel combination.
3. The two capacitors in the circuit c are in series combination.

The charging and discharging of a capacitor in an LC circuit is like an oscillatory motion. The period of these oscillations depends upon the values of the inductance and the capacitance in the circuit.

Formulae:

The angular frequency of an oscillation of a body, $\mathit{蝇}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{蟺}}{\mathbf{T}}$ (i)

The resonance frequency of an LC circuit, $\mathit{蝇}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{L}\mathbf{C}}}$ (ii)

The effective capacitance of a parallel circuit, ${\mathit{C}}_{\mathbf{e}\mathbf{f}\mathbf{f}}\mathbf{=}\underset{\mathbf{i}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{鈭�}}}{\mathit{C}}_{\mathbf{i}}$ (iii)

The effective capacitance of a series circuit, ${\mathit{C}}_{\mathbf{e}\mathbf{f}\mathbf{f}}\mathbf{=}\underset{\mathbf{i}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{鈭�}}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{i}}}$ (iv)

Consider circuit b. The two capacitors are connected in parallel combination. Thus, the effective capacitance of the circuit connection using equation (iii) is given as:

$C_b=C_1+C_2$

Since both the capacitors are identical, the effective capacitance becomes:

$C_b=2C$

Now, consider circuit c. The two capacitors are connected in series combination. Thus, the effective capacitance of the circuit using equation (iv) is given as:

$\frac{1}{C_c}=\frac{1}{C_1}+\frac{1}{C_2}$

Since both the capacitors are identical, thus the effective capacitance becomes

role="math" localid="1662748912891" $$\begin{gathered} \frac{1}{C_c}=\frac{C_1+C_2}{C_1{C}_2} \\ =\frac{2C}{C^2} \\ =\frac{2}{C} \\ {C}_c=\frac{C}{2} \end{gathered}$$

The period of oscillations is calculated by substituting equations (ii) in equation (i) as follows:

$T=2蟺\sqrt{LC}$

Thus, we see that$T鈭�\sqrt{LC}$

But, since the inductors in the three circuits are identical, $T鈭�\sqrt{C}$

Now, for circuit b, the effective capacitance is greatest among the three. So, its period is alsothegreatest. Thus, time for the capacitor to discharge fully, which is, will also be the greatest among the three.

For circuit a, the capacitance is C, which is smaller than that for circuit b. So, the time for the discharge will also be smaller.

For circuit c, the effective capacitance isthesmallest among the three. So, the time required for complete discharge will also bethesmallest.

Hence, the rank for the circuits is.$Circuitb>Circuita>Circuitc$