Question

The capacitors of capacitance 4 F, 6 F and 12 F are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases? (a) 2 : 3 (b) 11 : 1 (c) 1 : 11 (d) 1 : 3

Answer 1

Answer:

Given:

3 capacitors have been provided such as

• 4F
• 6F
• 12F

First they are connected in series and then in parallel.

To find:

Ratio of Equivalent Capacitance I'm both cases.

Calculation:

For series connection :

1/C eq. = 1/4 + 1/6 + 1/12

=> 1/C eq. = (3 + 2 + 1)/12

=> 1/ C eq. = 6/12 = ½

=> C eq. = 2 Farad.

For parallel connections:

C eq." = 4 + 6 + 12

=> C eq. = 22 Farad.

Required ratio :

C eq. : C eq."

= 2 : 22

= 1 : 11

So final answer is ( 1 : 11 )

Answer 2

$\huge{\orange{\underline{\red{\mathbb{GIVEN:-}}}}}$

3 CAPICITORS.

4F, 6F, 12F.

AND ARE CONNECTED SERIES AND PARALLEL.

$\huge{\orange{\underline{\red{\mathbb{ANSWER:-}}}}}$

CASE. 1)4F, 6F, 12F. CAPACITOR IS CONNECT IN PARALLEL.

$\frac{1}{c(eq)} \large\bold\red\rightarrow \frac{1}{4} + \frac{1}{6} + \frac{1}{12} \\ \frac{1}{c(eq)} \large\bold\red\rightarrow \frac{\frac{1}{4} \times 12 + \frac{1}{6} \times 12 + \frac{1}{12} \times 12}{12} \\ \frac{1}{c(eq)} \large\bold\red\rightarrow \frac{3 + 2 + 1}{12} \\ \frac{1}{c(eq)} \large\bold\red\rightarrow \frac{6}{12} \\ .°.c(eq.) \huge\bold\orange\rightarrow 2 \: farad.$

CASE. 2) 4F, 6F, 12F. CONNECT IN SERIES.

$\huge\bold\red\rightarrow$C(eq)'=4+6+12

$\huge\bold\red\rightarrow$(eq)'=22farad.

RATIO IS.

$\huge\bold\red\rightarrow$c(eq):c(eq)'

$\huge\bold\red\rightarrow$2:22

$\huge\bold\purple\rightarrow$1:11

.°. option c is the right answer.

$\huge{\orange{\underline{\red{\mathbb{THANKS.}}}}}$