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Question

G Google YouTubeMyuwindsor-Blackboard M Gmaill @ Pearson G chegg CIBC , pp n Facebook !! Micr osoft ", VMedia e AIR MILES-My Petro·SCENE Best BUY Assignment 3 Problem 24.53 Part A In the figure (Figure 1), Cl . G-8.5 F and G-G-G applied potential is Vab 220V What is the equivalent capacitance of the network between points a and b? Express your answer using two significant figures. 4.4 F . The Submit t Answer Part B Calculate the charge on capacitor C Express your answer using two significant figures. Submit Figure 1 of 1 Part C Calculate the potential difference across capacitor C Express your answer using two significant figures. C. 0 2 C.

Explanation / Answer

C3 and 4 are in series

C34 = 4.4 x 4.4/(4.4+4.4) = 2.2 uF

Now C23 is in parallel with C2

C34-2 = 2.2+ 4.4 = 6.6

Now C34-2; 1 and 5 are in series so

1/Ceq = 1/8.5 + 1/8.5 + 1/6.6 => Ceq = 2.6 uF

Hence, Ceq = 2.6 uF = 2.6 x 10^-6 F

b)Charge on C1 will be:

Q = Ceq V

Q = 2.6 x 10^-6 x 220 = 572 x 10^-6 C

Hence, Q1 = 572 x 10^-6 C = 5.72 x 10^-4 C

c)V1 = Q/C1

V1 = 572 x 10^-6/8.5 x 10^-6 = 67.3 V

Hence, V1 = 67.3 V

(d is solved after f)

e)V2 = 220 - 2 x 67.3 = 85.4 V

V2 = 85.4 V

(calculated in d)

f)Q2 + Q34 = 572 uC

V2 = V34

Q2/C2 = Q34/C34 => Q2/4.4 = Q34/2.2

Q2 = 2 Q34

2Q34 + Q34 = 572 => Q34 = 190.67 uC

Hence, Q3 = 190.67 x 10^-6 C = 1.91 x 10^-4 C

d)Q2 = 2 x 190.67 = 381.34 uC

Hence, Q2 = 381.34 x 10^-6 C = 3.81 x 10^-4 C

g)V3 + V4 = 85.4 V

Q3 = Q4

4.4 V3 = 4.4 V4 => V3 = V4

V3 = 85.4/2 = 42.7 V

Also, V4 = 42.7 V2

Hence, V3 = 42.7 V

g)V3 = 42.7 V

h)Q4 = 190.67 x 10^-6 C = 1.91 x 10^-4 C

(Since it is in series with C3)

i)V4 = 42.7 V

j)Q5 = Q1 = 572 x 10^-6 C = 5.72 x 10^-4 C

k)V5 = 572 x 10^-6/8.5 x 10^-6 = 67.3 V

Hence, V5 = 67.3 V

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