First slide

Capacitors and capacitance

Question

In the following circuit the resultant capacitance between A and B is $1 μ F$ . Then value of C is

Moderate

A

$\frac{32}{11} μ F$
B

$\frac{11}{32} μ F$
C

$\frac{23}{32} μ F$
D

$\frac{32}{23} μ F$

Solution

12 $μ F$ and 6 $μ F$ are in series and again are in parallel with4 $μ F$ therefore resultant of these three will be
$= \frac{12 \times 6}{12 + 6} + 4 = 4 + 4 = 8 μ F$
This equivalent system is in series with 1 $μ F$
Its equivalent capacitance $= \frac{8 \times 1}{8 + 1} = \frac{8}{9} μ F \to (1)$
Equivalent of 8 $μ F$ , 2 $μ F$ and 2 $μ F$
$= \frac{4 \times 8}{4 + 8} = \frac{8}{3} μ F \to (2)$
1 and 2 are in parallel and are in series with C
$∴ \frac{8}{9} + \frac{8}{3} = \frac{32}{9} C_{e q} = 1 = \frac{\frac{32}{9} \times C}{\frac{32}{9} + C} \Rightarrow C = \frac{32}{23} μ F$

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