Q.
In the given network capacitance, $C_{1} = 10\mu F,C_{2} = 5\mu F$ and $C_{3} = 4\mu F.$ What is the resultant capacitance between $A$ and $B$
A) $2.2\ \mu\ F$
B) $3.2\ \mu\ F$
C) $1.2\ \mu\ F$
D) $4.7\ \mu\ F$
Explanation
(b) $C = \frac{(C_{1} + C_{2}) \times C_{3}}{\left( C_{1} + C_{2} \right) + C_{3}} = \frac{(5 + 10) \times 4}{5 + 10 + 4} = \frac{60}{19} = 3.2\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.