Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 19 - DC Circuits - Search and Learn - Page 559: 1

Answer

$$\left(\begin{array}{r|r|r} {\rm Series} &{R_{eq}=R_1+R_2+...} & {C_{eq}=\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}+...\right)^{-1}} \\\\ \end{array}\right)$$ ---- $$\left(\begin{array}{r|r|r} { {\rm Parallel}} &{ R_{eq}=\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+...\right)^{-1}} & {C_{eq}=C_1+C_2+...} \\ \end{array}\right)$$

Work Step by Step

Comparing the formulas as the author asked: $$\left(\begin{array}{r|r|r} {\rm Series} &{R_{eq}=R_1+R_2+...} & {C_{eq}=\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}+...\right)^{-1}} \\\\ \end{array}\right)$$ ---- $$\left(\begin{array}{r|r|r} { {\rm Parallel}} &{ R_{eq}=\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+...\right)^{-1}} & {C_{eq}=C_1+C_2+...} \\ \end{array}\right)$$ When we connect resistors in series, the net resistance we got is the sum of all of these in series resistors. The same current passes through each one of them. So $$\boxed{R_{eq}=R_1+R_2+...}$$ And when we connect the resistors in parallel, they will have the same voltage, but not the same resistors, so the net resistance is given by $$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+...$$ which is as same as; $$ \boxed{R_{eq}=\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+...\right)^{-1}}$$ When we connect capacitors in series, the net capcitance we got is then given by $$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+...$$ which is as same as; $$\boxed{C_{eq}=\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}+...\right)^{-1}}$$ But when we connect them in parallel the net capactiance is given by $$\boxed{C_{eq}=C_1+C_2+...}$$ Now we need to find the relation between the resistance and the voltage; $$V=IR$$ Thus, $$R=\dfrac{V}{I}$$ It is obvious now that the resistance is proportional to the voltage $R\propto V$ While the capacitance is related to the vooltage by $$C=\dfrac{Q}{V}$$ which means that it is inversely proportional to the voltage $C\propto \dfrac{1}{V}$
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