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ORELA Physics (308) Practice Tests & Test Prep by Exam Edge - Free Test


Our free ORELA Physics (308) Practice Test was created by experienced educators who designed them to align with the official Oregon Educator Licensure Assessments content guidelines. They were built to accurately mirror the real exam's structure, coverage of topics, difficulty, and types of questions.

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ORELA Physics - Free Test Sample Questions

The effective capacitance of two capacitors C1 and C2 ( where C2>C1), are connected in parallel.   It is 25/6 times the effective capacitance when they are connected in series. What is the ratio of C2/C1?

Correct Answer:
3/2
to solve the problem involving the ratio of two capacitors c1 and c2 when connected in parallel and in series, we start by using the given condition that the effective capacitance when the capacitors are connected in parallel is \( \frac{25}{6} \) times the effective capacitance when they are connected in series.

let's denote the capacitances of the two capacitors as c1 and c2, where c2 > c1. the effective capacitance \( c_{\text{parallel}} \) when capacitors are connected in parallel is given by: \[ c_{\text{parallel}} = c1 + c2 \]

when the capacitors are connected in series, the effective capacitance \( c_{\text{series}} \) is given by: \[ \frac{1}{c_{\text{series}}} = \frac{1}{c1} + \frac{1}{c2} \] \[ c_{\text{series}} = \frac{c1 \times c2}{c1 + c2} \]

according to the problem, the effective capacitance in parallel is \( \frac{25}{6} \) times the effective capacitance in series: \[ c1 + c2 = \frac{25}{6} \times \frac{c1 \times c2}{c1 + c2} \] to simplify, multiply both sides by \( c1 + c2 \): \[ (c1 + c2)^2 = \frac{25}{6} \times c1 \times c2 \] \[ 6(c1 + c2)^2 = 25c1c2 \]

expanding and rearranging gives: \[ 6c1^2 + 12c1c2 + 6c2^2 = 25c1c2 \] \[ 6c1^2 + 6c2^2 - 13c1c2 = 0 \]

let's assume \( c2 = xc1 \) (where x > 1 since c2 > c1). substituting gives: \[ 6c1^2 + 6(xc1)^2 - 13(xc1)c1 = 0 \] \[ 6 + 6x^2 - 13x = 0 \]

solving this quadratic equation: \[ x^2 - \frac{13}{6}x + 1 = 0 \] using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where a = 1, b = -\(\frac{13}{6}\), and c = 1: \[ x = \frac{\frac{13}{6} \pm \sqrt{(\frac{13}{6})^2 - 4}}{2} \] \[ x = \frac{\frac{13}{6} \pm \frac{5}{6}}{2} \] \[ x = \frac{3}{2} \text{ or } x = \frac{2}{3} \]

since c2 > c1, we discard \( x = \frac{2}{3} \) and accept \( x = \frac{3}{2} \). therefore, \( \frac{c2}{c1} = \frac{3}{2} \). this indicates that the capacitance of c2 is 1.5 times that of c1.