SSC JE Electrical Engineering 23 Jan 2018 Evening Paper

Question : 107 of 200

Marks: +1, -0

What will the equivalent capacitance (in mF) of three capacitor s connected in a series having capacitance of 0.04 mF, 0.08 mF and 0.02 respectively?

Solution:

We know

q = cv
in series charge is constant and voltage depend on capacitor

q=v_1=q/c_1

v_2=q/c_2

v_3=q/c_3

v = v_1 + v_2 + v_3

q/c_\eq=q/c_1+q/c_2+q/c_3

=1/0.04+1/0.08+1/0.02

= 0.011 mF

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