Question

n capacitors of equal capacitance arw connected to give maximum and minimum capacitance.Find the ratio of maximum to minimum capacitance?​

Answer 1

Answer:the maximum capacitance will be in the case of parallel combination and the minimum capacitance will be in the case of series combination. let us consider all the capacitors were of capacitance C. now the required ratio of maximum to minimum capacitance is = Ceq/C'eq = N2

Explanation:

Answer 2

RATIO WILL BE n² : 1

Given:

• n capacitors (equal capacitance) are connected to give maximum and minimum capacitance.

To find:

• Ratio of max to min net capacitance?

Calculation:

Equivalent capacitance is maximum when the capacitors are connected in parallel combination such that :

$C_{max} = C + C + ... + \: n \: times$

$\implies C_{max} =n C$

Equivalent capacitance is minimum when the capacitors are connected in series combination such that :

$\dfrac{1}{C_{min}} = \dfrac{1}{C} + \dfrac{1}{C} + ... n \:t imes$

$\implies \dfrac{1}{C_{min}} = \dfrac{n}{C}$

$\implies C_{min} = \dfrac{C}{n}$

• So, required ratio will be :

$C_{max} : C_{min} = {n}^{2} : 1$