Question

how will you arrange 3 capacitors each having the capacity of 2uf, to get a capacitor of capacity 3uf?​

Answer 1

Answer:

We arrange one of the 2μF capacitors in parallel to the other two capacitors which will be arranged in series

Explanation:

The number of capacitors in the network = 3

The capacitance of the each capacitor, C₁, C₂, and C₃ = 2 μF

The sum of capacitors in series = The inverse of the sum of the reciprocals of the capacitances of the capacitor

$C_{total \ series} = \dfrac{1}{\dfrac{1}{C_a} +\dfrac{1}{C_b} }$

The sum of capacitances of capacitors arranged in parallel = The sum of the individual capacitances in parallel

$C_{total \ parallel}$ = Cₐ + C$_b$

By placing two of the capacitors in series, and the third in parallel to the first two, we get;

$C_{total} = C_{total \ series} + C_{total \ parallel}$

$C_{total \ series} =C_{1 \, and \, 2} = \dfrac{1}{\dfrac{1}{2} +\dfrac{1}{2} } = \dfrac{1}{1} = 1$

$C_{total \ parallel}$ = $C_{1 \, and \, 2}$ + C₃

∴ $C_{total \ parallel}$ = 1 μF + 2μF= 3 μF

Therefore, to get a capacitor of capacity of 3 μF from 3 capacitors of

2 μF, one of the capacitors is arranged in parallel across the other two capacitors arranged in series.