1. The problem statement, all variables and given/known data See capacitor design in attachment. Calculate the equivalent capacitance of the circuit shown in the diagram above; where C1 = 4.30 uF, C2 = 7.15 uF, C3 = 2.25 uF, and C4 = 3.75 uF. How much charge is drawn from the battery? 2. Relevant equations Parallel C = c1 +c2... Series C = 1/c1 +1/c2 ... C=Q/V 3. The attempt at a solution For the first question, I drew the capacitor in a form that is more familiar than the triangle. I made C2 and C3 parallel to each other, and C4 in between them in the circuit. From there, I added C2 and C3 together, and since they are in parallel, I got the answer to be 9.4 uF. From there all the circuits are in series so I added 1/4.30 uF + 1/94 uF + 1/3.75 uF and got 1/.6056 or 1.65 uF. This however, is not the correct answer. For the second question, I used C=Q/V, however, having C wrong, my answer was wrong.