Q. Three capacitors each of capacitance 1$\ \mu$F are connected in parallel. To the combination, a fourth capacitor of capacitance 1 $\mu$F is connected in series. The resultant capacitance of the system is
A) $4\ \mu$F
B) $(4/3)\mu$F
C) $2\ \mu$F
D) $(3/4)\mu$F
Explanation
(d) $C_{p} = 1 + 1 + 1 = 3\mu\ F$
$\frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{1} = \frac{4}{3}$
$\therefore C_{s} = \frac{3}{4}\mu\ F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.