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**What is the equivalent capcitance of this circuit! -- Figured it out.**

I have a Physics final exam in 2 hours and I'm re-reviewing things....

This problem DOESN'T make any sense!!

17. If C

17. If C

_{1}= 2 F, C_{2}= 3 F, C_{3}= 5 F, C_{4}= 6 F, the equivalent capacitance of the circuit is:**A. 8.5 F**is supposed to be the answer - that makes NO sense

B. 16 F

C. 5.5 F

D. 10 F

I didn't get ANY of those numbers...

This is what I did.

1) Determine PARALLEL capacitance of C

_{1}and C

_{2}

C

_{eq}= C

_{1}+ C

_{2}

C

_{eq}= 2 +3

C

_{eq}= 5

2) Determine the SERIES capacitance of the whole circuit now

1/C

_{eq}= 1/5 + 1/C

_{3}+ 1/C

_{4}

1/C

_{eq}= 1/5 + 1/5+ 1/6

1/C

_{eq}= 0.567

C

_{eq}= 1/0.567

C

_{eq}= 1.763 :(:(:(:(!!!!!

Edit:

Oh wait nvm.

I figured it out.

the upper arm (C net) is- 1/C net = 1/5 + 1/C3 = 1/5 + 1/5 =2/5

=>C net = 5/2 = 2.5

now 2.5 and 6 in parallel- 2.5+6 = 8.5

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