A capacitance C 1 = 10.1 ?F is connected in series with a capacitance C 2 = 2.5
ID: 2140047 • Letter: A
Question
A capacitance C1 = 10.1 ?F is connected in series with a capacitance C2 = 2.5 ?F, and a potential difference of 150 V is applied across the pair.
(A) Calculate the equivalent capacitance.
(B) What is the charge on C1?
(C) What is the charge on C2?
(D) What is the potential difference across C1?
(E) What is the potential difference across C2?
(F) Repeat for the same two capacitors but with them now connected in parallel. Calculate the equivalent capacitance.
(G) What is the charge on C1?
(H) What is the charge on C2?
(I) What is the potential difference across C1?
(J)What is the potential difference across C2?
Explanation / Answer
A)Cnet=C1*C2/(C1+C2)=2.003 uF
B)charge on C1 is Q1=Cnet*V=300.45uC...
C)charge on C2 is Q2=Cnet*V=300.45uC...
D)p.d=V1=Q1/C1= 29.74V..
E)p.d=V2=Q2/C2=120.18 V...
F)Cnet=C1+C2=12.6uF....
G)Q1=C1*V=10.1uF*150=1515 uC...
H)Q2=C2*V=375uC..
i)V1=150 V..
J)150 V
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