Q.
In the arrangement of capacitors shown in figure, each capacitor is of 9$\ \mu$F, Then the equivalent capacitance between in points A and B is
A) 9$\ \mu$F
B) 18$\ \mu$F
C) 4.5$\ \mu$F
D) 15$\ \mu$F
Explanation
(d) The arrangement can be redrawn as shown in the adjoining figure.
$C_{13} = C_{1} + C_{3} = 9 + 9 = 18\mu F$
$C_{2 - 13} = \frac{C_{2} \times C_{13}}{C_{2} + C_{13}\ \ \ } = \frac{9\mu F \times 18\mu F}{(9 + 18)\mu F} = 6\mu F$
$\therefore C = C_{2 - 13} + C_{4} = 6\mu F \times 9\mu F = 15\mu F$.
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.