Question

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

• A. $\frac{4}{7} pF$
• B. 1 PE
• C. $\frac{7}{4} pF$
• D. 2 pF

Answer 1

Answer: (A)

$\frac{4}{7} pF$

Answer 2

The minimum value of effective capacitance is get in series combination. Now, the equivalent capacitance.
$\frac{1}{C_{\text {eq }}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}$
$\frac{1}{C_{ eq }} =\frac{4+2+1}{4}$
$\frac{1}{C_{ eq }} =\frac{7}{4}$
$C_{ eq } =\frac{4}{7} \,pF$