The minimum value of effective capacitance th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The effective capacitance is minimum when all capacitors are connected in series.For capacitors connected in series,the equivalent capacitance $C_

Vedclass · Accepted Answer

$4/7 \ pF$

Vedclass · Answer

(A) The effective capacitance is minimum when all capacitors are connected in series.For capacitors connected in series,the equivalent capacitance $C_{eff}$ is given by the formula:$\frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$Given $C_1 = 1 \ pF$,$C_2 = 2 \ pF$,and $C_3 = 4 \ pF$.Substituting the values:$\frac{1}{C_{eff}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4}$$\frac{1}{C_{eff}} = \frac{4 + 2 + 1}{4} = \frac{7}{4} \ pF^{-1}$Therefore,$C_{eff} = \frac{4}{7} \ pF$.

The minimum value of effective capacitance that can be obtained by combining $3$ capacitors of capacitances $1 \ pF, 2 \ pF$ and $4 \ pF$ is

Similar Questions

All six capacitors shown are identical. Each can withstand a maximum of $200\, V$ between its terminals. The maximum voltage that can be safely applied between $A$ and $B$ is.....$V$

Two capacitors of equal capacity are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be

Capacitors of $C, 2C, 4C, \dots, \infty$ are connected in parallel. What will be their equivalent capacitance?

Three capacitors each of capacity $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by:

In the given figure,the potential at point $A$ is $900 \ V$ and point $B$ is earthed. What will be the potential at point $P$ (in $V$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module