Three capacitors of capacitances 1 µF, 2µF and 4μF are connected first in a series combination and then in a parallel combination. The ratio their equivalent capacitances will be

2 : 49

49 : 2

4 : 49

49 : 4

Solution

4 : 49

Capacitors are connected together in series when they are daisy chained together in single

In series combination

$\frac{1}{C_{1}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4}$

= $\frac{4 + 2 + 1}{4}$

$\frac{1}{C_{1}} = \frac{7}{4}$

⇒ $C_{1} = \frac{4}{7} μF$

Capacitors are connected together in parallel when both of its terminals are connected to each terminal of another capacitor.

Parallel capacitors equation

C_T = C₁ + C₂ + C₃ + ...

In parallel combination,

C₁ = 1 + 2 + 4

= 7 μF

∴ $\frac{C_{1}}{C_{2}} = \frac{\frac{4}{7}}{7}$

$\frac{C_{1}}{C_{2}} = \frac{4}{49}$

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