For the network in Figure 2, determine the equivalent capacitance across termina
ID: 1925652 • Letter: F
Question
For the network in Figure 2, determine the equivalent capacitance across terminals a and b. Assume that v1= 2.0 V, v2= 6.0 V, v3 = 10.0 V, v5 = 10.0 V, v6 = 3.0 V, C1= 2.0 F, C2 = 2.0 F, C3 = 8.0 F, C4 = 9.0 F, C5 = 6.0 F, C6 = 10.0 F, C7 = 10.0 F, and C8 = 2.0 F.
My attempt at a solution:
EQUATIONS USED:
In series: 1/C=1/C1+1/C2+...+1/Cn
In parallel: C=C1+C2+...+Cn
First I summed up C1 and C3 inversely, yielding an answer of 1.6 F
Next I added C1 and C3 with C4 (i.e. [C1 + C3] + C4) and got a result of 10.6 F
Then I added C6 and C8 inversely and got 1.667 F, and added C5 and C7 inversely and got an answer of 3.7495 F
Next I added those together ([C6 + C8]+[C5 + C7]) and got an answer of 5.4165 F
Finally I added (1/C2)+(1/10.6 F)+(1/5.4165 F) and got the final answer of 1.284 F...WHICH IS WRONG!!!
Where did I mess up?!
Explanation / Answer
1/C = (1/C1) + (1/C3), which is 1/C = 2, then C = (8/5)F, this will give you 1.6F
Then you can add C and C4 in parallel which is C = C + C4 = 1.6F + 9F = 10.6F
Adding C6 and C8 in series you get: 1.667F
Adding C5 and C7 in series you get: 3.750F
Adding those two in parallel you get 5.417F
Now, you add 10.6F (which you got from the left side) and 5.417F in series
1/C = (1/10.6F)+(5.417F)
C = 3.585F
Finally you add that to C to C2 in parallel: 3.585F + 2F
The Ceq (Equivalent Capacitance) will be equal to 5.585F