Three capacitors each of capacitance 6 PF are connected in series. What is the
total capacitance of the combination?
Answers
Answer:
Here, C
1 =C
2=C
3 =9pF=9×10 −12 F
V=120volt
(a) Total capacitance of the series combination in given by
C1 = C 1
1+ C 2
1+ C 3
1 =3× 9×10 −12
1 = 3×10 −121
∴C=3×10 −12
F=3pF
(b) Let q be the charge on each capacitor. Then, sum of the potential differences across the plates of the three capacitors must be equal to 120 V i.e.,
V
1+V 2
=120 or C 1
q+C2
q+C 3
q=120or 9×10 −12
q+9×10 −12
9×10−12
3q =120 or q=360×10 −12 C
Therefore, potential difference across a capacitor
= capacitance
q= 9×10 −12
360×10 −12
S=40V
Alternative method. Since the three capacitors are of the same capacitance, potential difference across each capacitor
= 3
120
=40V.
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Answer:
Capacitors in series combination :
Explanation:
when capacitors are connected inserted combination their equivalent capacitance is given by the reciprocal of the equipment capacitance in series is given by the sun of the reciprocals of the given capacitors.
here given,
number of capacitors n = 3 ,
capacitance C1, C2 and C3 all identical and equal to 6 PF,
equivalent capacitance :
C = 2 PF
thus, the equivalent capacitance of the combination is 2 PF.