Question

problem 7: (6% of assignment value) three capacitors are connected as shown, where c = 0.015 f. part (a) input an expression for the equivalent capacitance between points a and b, in terms of c. hints: 4% deduction per hint. hints remaining: 2 feedback: 5% deduction per feedback. grade summary deductions potential submissions attempt(s) remaining: 3 4% deduction per attempt detailed view

Explanation:

Step1: Identify capacitor - combination rules

For capacitors in parallel, \(C_{eq - parallel}=C_1 + C_2+\cdots\), and for capacitors in series, \(\frac{1}{C_{eq - series}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots\). First, note that \(3C\) and \(2C\) are in parallel.

Step2: Calculate the equivalent of \(3C\) and \(2C\) in parallel

\(C_{1}=3C + 2C=5C\) (using the parallel - capacitor formula).

Step3: Calculate the equivalent of \(C_{1}\) and \(5C\) in series

Since \(C_{1} = 5C\) and the other capacitor is \(5C\), and they are in series, \(\frac{1}{C_{eq}}=\frac{1}{5C}+\frac{1}{5C}=\frac{2}{5C}\). Then, \(C_{eq}=\frac{5C}{2}\).

Answer:

\(\frac{5C}{2}\)