Consider capacitors $$c _ { 1 } = 1 \mathrm {~F} , c _ { 2 } = 4 \mathrm {~F} , c _ { 3 } = 9 \mathrm {~F} , c _ { 4 } = 16 \mathrm {~F} , \ldots , c _ { n } = n ^ { 2 } \mathrm {~F}$$ with $n \rightarrow \infty$ . If these infinite number of capacitors are connected in series, the effective capacitance as seen by the source is
A) 0 F.
B) $\infty \mathrm { F } .$
C) $6 / \pi ^ { 2 } \mathrm {~F} .$
D) 3/7 F.

Question

Quizplus · Accepted Answer

The effective capacitance, $C_{\text{eff}}$, of capacitors connected in series is given by the reciprocal of the sum of the reciprocals of their individual capacitances. For the given series of capacitors, where each capacitor has a capacitance of $n^2$ farads, the formula for the effective capacitance becomes:$$ \frac{1}{C_{\text{eff}}} = \sum_{n=1}^{\infty} \frac{1}{n^2} $$This sum is a well-known series in mathematics, the Basel problem, and its sum converges to $\frac{\pi^2}{6}$. Therefore, the effective capacitance is:$$ C_{\text{eff}} = \frac{1}{\pi^2/6} = \frac{6}{\pi^2} \mathrm{~F} $$

Consider Capacitors With n→∞n \rightarrow \inftyn→∞ If These Infinite Number of Capacitors Are Connected in {~F}

Consider Capacitors With $n \rightarrow \infty$ If These Infinite Number of Capacitors Are Connected in {~F}