Question

(i) Find the equivalent capacitance of the infinite ladder shown in the figure between the terminals A and B. (ii) If now each capacitor is replaced by a capacitor which is double in capacitance, then reevaluate the question.

Answer 1

Hint: For both cases, assume that a capacitor similar to the equivalent capacitance is connected across the second parallel branch of the circuit, since the equivalent capacitance between A and B remains unchanged from such an assumption. Then arrive at a quadratic equation for the equivalent capacitance and solve it. Eliminate the negative solution since capacitance cannot have a negative value to arrive at a positive value for the equivalent capacitance, which is what is required.

Formula used:
$C_{series} = \dfrac{C_1 \times C_2}{C_1+C_2}$
$C_{parallel} = C_1 + C+2$

Complete step-by-step answer:
(i)Let us assume that the equivalent capacitance between A and B of the infinite ladder shown is C.
Assuming that a capacitance equivalent to C is connected in the circuit as shown in the figure below. Since it is one of the parallel branches across A and B, the equivalent capacitance between A and B still remains unchanged.
       

We now obtain an expression for the equivalent capacitance, say $C_{eq}$ across A and B from the circuit shown in the above figure.
Since the capacitances $6\mu F$ and$\;C$ are connected in series to each other, the cumulative capacitance, say $C^{\prime}$ of the two will be:
$C^{\prime} = \dfrac{C_1 \times C_2}{C_1 +C_2} = \dfrac{6 \times C}{6+C} = \dfrac{6C}{6+C}$
Now, the $8\mu F$ capacitance is parallel to $C^{\prime}$. Therefore, the equivalent capacitance $C_{eq}$ between A and B will be:
$C_{eq} = 8 + C^{\prime} = 8 + \dfrac{6C}{6+C} = \dfrac{8(6+C)+6C}{6+C} = \dfrac{48 + 14C}{6+C}$
But we’ve established that $C_{eq} = C$, therefore, the above equation becomes:
$C = \dfrac{48+14C}{6+C}$
$\Rightarrow C(6+C) = 48+14C$
$\Rightarrow C^2 + 6C – 14C -48 = 0$
$\Rightarrow C^2 -8C - 48 =0$
$\Rightarrow C^2 -12C + 4C - 48 = 0$
$\Rightarrow C(C-12) + 4(C - 12) = 0$
$\Rightarrow (C+4)(C-12) = 0$
$\Rightarrow C = -4\mu F$ or $C = 12\mu F$
Since$\;C$ cannot be negative, the appropriate choice would be $C = 12\mu F$
Therefore, the equivalent capacitance between terminals A and B will be $12\mu F$.
(ii) Given that each capacitor is replaced by a capacitor which is double in capacitance, as shown in the figure below.

        

Following the same procedure as before, we get that the equivalent capacitance between A and B will be:
${C^{\prime}}_{eq} = \dfrac{12 \times C}{12+C} + 16 = \dfrac{12C + 16(12+C)}{12+C} = \dfrac{12C + 192 + 16C}{12+C}$
Plugging in ${C^{prime}} _{eq} = C$, the above equation becomes:
$C = \dfrac{28C + 192}{12+C}$
$\Rightarrow C (12+C) = 28C +192 $
$\Rightarrow C^2 +12C -28C -192 = 0$
$\Rightarrow C^2 -16C -192 = 0$
$\Rightarrow C^2 -24C + 8C -192 = 0$
$\Rightarrow C(C-24) + 8(C-24) = 0$
$\Rightarrow (C+8)(C-24) = 0$
$\Rightarrow C = -8\mu F$ or $C = 24\mu F$
Since$\;C$ cannot be negative, the appropriate choice would be $C = 24\mu F$
Therefore, when the capacitances are doubled the equivalent capacitance between terminals A and B will be $24\mu F$, which is twice the equivalent capacitance between them from the previous case.

Note: Remember that for capacitors in parallel, the net capacitance is the additive sum of individual capacitances, whereas for capacitors in parallel, the reciprocal of the net capacitance is the sum of the reciprocals of individual capacitances. The largest effective capacitance is obtained by connecting the capacitors in parallel, whereas the smallest effective capacitance is obtained by connecting the capacitors in series.
Do not get this confused with resistors, where for resistors in parallel, the reciprocal of the net resistance is the sum of the reciprocals of individual resistances resulting in the smallest effective resistance, whereas for resistors is series, the net resistance is the additive sum of individual resistances, resulting in the largest effective resistance.