Question

Find the equivalent capacitance between A and B for this infinite circuit of capacitors.

• A. $\frac{(\sqrt{5}-1)C}{2}$
• B. $\frac{(\sqrt{5}+1)C}{2}$
• C. $\frac{(1-\sqrt{5})C}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{(\sqrt{5}-1)C}{2}$

You’ll see a lot of questions like these and they’re all solved in the same simple manner.
Consider the infinite combination as a “collection”. If I add one more unit to it, it should be fine right?
Because $\infty +1=\infty$. Now the unit here is,

So, if add this to the circuit, we simply have the same $C_{eq}$

Now the circuit is pretty simple isn’t it?
We have the same $C_{eq}$ across A and B.
$C_{eq}=C+\left(\frac{1}{\frac{1}{C}+\frac{1}{C_{eq}}}\right)$
$C_{eq}=C+\frac{C{C}_{eq}}{C+C_{eq}}$
$C{C}_{eq}+C_{eq}^2=C^2+C{C}_{eq}+C{C}_{eq}$
$C_{eq}^2+C{C}_{eq}-C^2=0$
$C_{eq}=\frac{-C\pm \sqrt{C^2+4C^2}}{2}$
$C_{eq}=\frac{-C\pm \sqrt{5C}}{2}$
Clearly Capacitance cannot be negative. So, taking only the positive root,
$C_{eq}=\frac{(\sqrt{5}-1)C}{2}$


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