Question

Given four capacitors each of capacity $12\mu F$ . To get a capacity of $9\mu F$ , what combination can be used:
(A) All in series
(B) All in parallel
(C) 3 in parallel and 1 series with them
(D) 2 in parallel and 2 in series

Answer 1

Hint: In order to answer this question, first we will write the formula for equivalent capacitance for both the cases of capacitance. And then we will find the equivalent capacitance by putting the given values.

Complete step by step solution:
Formula for equivalent capacitance in parallel:
${C_{eq}} = {C_1} + {C_2}$
Again, the formula for equivalent capacitance in series:
$\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Therefore, for 3 capacitance, $9\mu F$ each in parallel and 1 series, the equivalent capacitance:-
$
  \therefore \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{36}} + \dfrac{1}{{12}} \\
   \Rightarrow {C_{eq}} = \dfrac{{36 \times 12}}{{36 + 12}} \\
   \Rightarrow {C_{eq}} = 9\mu F \\
 $
Hence, the correct option is (C) 3 in parallel and 1 series with them.

Additional Information:- There are two ways to link capacitors: in series and in parallel. When capacitors are connected in series, they are connected one after the other in a chain. The capacitance is lower in series. The capacitors are said to be connected in parallel when they are connected between two common locations.

Note: The total capacitance for parallel capacitors is computed using the formula \[capacitance = Capacitance + Capacitance\] . Capacitance \[\left( {C1} \right)\] and Capacitance \[\left( {C2} \right)\] are required to compute Equivalent Capacitance for Capacitors in Parallel.