Question

two 33uf capacitors are connected in series with each other what will there combined capacitance be in faradas

Answer 1

Answer:

16.5

Explanation:

Given ; C1 = 33 F

C2 = 33 F

connected in series

resultant ;

1/C = 1/C1 + 1/C2

Putting the value

1/33 +1/33

2/33

resultant ; C = 33/2 = 16.5 F

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Answer 2

Answer:

The combined capacitance of the capacitor connected in series is 16.5 F

Explanation:

Given: Two 33uf capacitors are connected in series with each other

To find: The combined capacitance in faradas.

Solution

Two or more capacitors in series will always have equal amounts of coulomb charge across their plates

Two capacitors are connected in series

Therefore

C1 =33uf

C2 = 33 uf

As we know that

V= Q/C

$V =q(\frac{1}{C_{1} } +\frac{1}{C_{2} } )$

Effective capacitance,

$\frac{Q}{C_{eff} } =q(\frac{1}{C_{1} } +\frac{1}{C_{2} } )$

$\frac{1}{C_{eff} } =(\frac{1}{C_{1} } +\frac{1}{C_{2} } )$

The combined capacitance formula from the above derivation

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

⇒ $\frac{1}{C} = \frac{1}{33} + \frac{1}{33}$

⇒ $\frac{1}{C} = \frac{2}{33}$

Taking reciprocal

$C = \frac{33}{2}$

C = 16.5  F

Final answer:

The combined capacitance of the capacitor connected in series is 16.5 F

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