In the figure below, a potential difference V = 140 V is applied across a capaci
ID: 1448609 • Letter: I
Question
In the figure below, a potential difference V = 140 V is applied across a capacitor arrangement with capacitances C1 = 11.0 F, C2 = 7.00 F, and C3 = 2.00 F. Find the following values.
(a) the charge on capacitor 3
C
(b) the potential difference across capacitor 3
V
(c) the stored energy for capacitor 3
J
(d) the charge on capacitor 1
C
(e) the potential difference across capacitor 1
V
(f) the stored energy for capacitor 1
J
(g) the charge on capacitor 2
C
(h) the potential difference across capacitor 2
V
(i) the stored energy for capacitor 2
J
In the figure below, C1 = 10.5 F, C2 = 17.5 F, and C3 = 26.0 F. If no capacitor can withstand a potential difference of more than 100 V without failure, what are the following.
(a) the magnitude of the maximum potential difference that can exist between points A and B
V
(b) the maximum energy that can be stored in the three-capacitor arrangement
J
A coaxial cable used in a transmission line has an inner radius of 0.19 mm and an outer radius of 0.74 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with polystyrene. (Also assume that the outer conductor is infinitesimally thin.)
pF/m
Explanation / Answer
a) charge on capacitor 3=CV=140*2=280C
b) potential difference across capacitor 3=V=140V
c)the stored energy for capacitor 3=0.5CV2=0.5*2*140*140=19600J=0.0196J
d)the charge on capacitor 1=C1(V-X)
where V across capacitor 1 can be calculated as
(X-V)C1+(X-0)C2=0
X=VC1/(C1+C2)=140*11/18=85.55V
the charge on capacitor 1=C1(V-X)=11*(140-85.5)=599.5C
e) the potential difference across capacitor 1=(V-X)=(140-85.5)=54.5V
f) the stored energy for capacitor 1=0.5CV2=0.5*11*54.5*54.5=16336.375J=0.0163J
g)the charge on capacitor 2=C1(X-0)=7*(85.5)=578.5C
h)the potential difference across capacitor 2=(X-0)=85.5V
i) the stored energy for capacitor 2=0.5CV2=0.5*7*85.5*85.5=25585.875J=0.0256J
NEXT QUESTION
a)Net Capacitance=[(10.5+17.5)/(10.5*17.5)+(1/26)]-1=5.3157F
let the magnitude of the maximum potential difference that can exist between points A and B be V.
Net charge flowing=CV=5.3157V
Potential difference across capacitor 1=Q/C=5.3157V/10.5=0.506V
Potential difference across capacitor 2=Q/C=5.3157V/17.5=0.3037V
Potential difference across capacitor 3=Q/C=5.3157V/26=0.2044V
maximum among all potential difference across capacitors=100=0.506V
Vmax=197.63Volts
b)Maximum energy that can be stored=0.5CV2=0.5*5.3157*197.63*197.63=103809.287078J=0.10381J
NEXT QUESTION
C/L=20K/ln(b/a)
C/L=2*3.14*(8.85*10^-12)*2.6/ln(0.74/0.19)=106.288pF/m