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Expert-verified Found in: Page 795 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # After the switch in Fig. 27-15 is closed on point a, there is current i through resistance R. Figure 27-23 gives that current for four sets of values of R and capacitance C: (1) ${{\mathbf{R}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{0}}}$, (2) ${\mathbf{2}}{{\mathbf{R}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{0}}}$, (3) ${{\mathbf{R}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{2}}{{\mathbf{C}}}_{{\mathbf{0}}}$, (4) ${\mathbf{2}}{{\mathbf{R}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{2}}{{\mathbf{C}}}_{{\mathbf{0}}}$. Which set goes with which curve? Nature of graph for the given conditions is as follows:

1-(d)

2 -(a)

3-(c)

4- (b)

See the step by step solution

## Step 1: Given

${\mathrm{R}}_{0}\mathrm{and}{\mathrm{C}}_{0}\phantom{\rule{0ex}{0ex}}2{\mathrm{R}}_{0}\mathrm{and}{\mathrm{C}}_{0}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{0}\mathrm{and}2{\mathrm{C}}_{0}\phantom{\rule{0ex}{0ex}}2{\mathrm{R}}_{0}\mathrm{and}2{\mathrm{C}}_{0}\phantom{\rule{0ex}{0ex}}$

## Step 2: Determining the concept

Use the formula for charge in terms of capacitor and voltage and then find the current from charge, resistance, and capacitor. Further, we can use time constant values. Higher the values of RC, lower will be state of decreasing of current.

Formulae are as follow:

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\mathrm{IR}\phantom{\rule{0ex}{0ex}}\mathrm{i}=-\frac{\mathrm{q}}{\mathrm{RC}}*{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{RC}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Where, C is capacitance, V is potential difference, q is charge, I is current, V is voltage, R is resistance, t is time.

## Step 3: Determining the nature of graph for the given conditions

$\mathrm{Case}1\right){\mathrm{R}}_{0}\mathrm{and}{\mathrm{C}}_{0}$

First find the charge as follows:

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}\mathrm{q}={\mathrm{C}}_{0}{\mathrm{V}}_{0}$

Current is given as follows:

${\mathrm{I}}_{0}=\frac{{\mathrm{q}}_{0}}{{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{C}}_{0}{\mathrm{V}}_{0}}{{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{V}}_{0}}{{\mathrm{R}}_{0}}$

The time constant for this case will be ,

$\mathrm{t}={\mathrm{R}}_{0}{\mathrm{C}}_{0}$

$\mathrm{Case}2\right)2{\mathrm{R}}_{0}\mathrm{and}{\mathrm{C}}_{0}$

First find the charge as follows:

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{0}={\mathrm{C}}_{0}{\mathrm{V}}_{0}$

Current is given as follows:

${\mathrm{I}}_{0}=\frac{{\mathrm{q}}_{0}}{2{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{C}}_{0}{\mathrm{V}}_{0}}{2{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{V}}_{0}}{2{\mathrm{R}}_{0}}$

The time constant is,

$\mathrm{t}=\mathrm{RC}=2{\mathrm{R}}_{0}{\mathrm{C}}_{0}$

$\mathrm{Case}3\right){\mathrm{R}}_{0}\mathrm{and}2{\mathrm{C}}_{0}$

First find the charge as follows:

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{0}=2{\mathrm{C}}_{0}{\mathrm{V}}_{0}$

Current is given as follows:

${\mathrm{I}}_{0}=\frac{{\mathrm{q}}_{0}}{2{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{2{\mathrm{C}}_{0}{\mathrm{V}}_{0}}{2{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{V}}_{0}}{{\mathrm{R}}_{0}}$

The time constant for this case is,

$\mathrm{t}=\mathrm{RC}=2{\mathrm{R}}_{0}{\mathrm{C}}_{0}$

$\mathrm{Case}4\right)2{\mathrm{R}}_{0}\mathrm{and}2{\mathrm{C}}_{0}$

First find the charge as follows:

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}\mathrm{q}=2{\mathrm{C}}_{0}{\mathrm{V}}_{0}$

Current is given as follows:

${\mathrm{I}}_{0}=\frac{{\mathrm{q}}_{0}}{4{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{2{\mathrm{C}}_{0}{\mathrm{V}}_{0}}{4{\mathrm{R}}_{0}{\mathrm{C}}_{0}}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}_{0}=\frac{{\mathrm{V}}_{0}}{2{\mathrm{R}}_{0}}$

The time constant for this case is,

$\mathrm{t}=\mathrm{RC}=4{\mathrm{R}}_{0}{\mathrm{C}}_{0}$

Now, if considered all cases, it can be seen that the case 1 and 3 have higher current values.

So, between these two, the value of time constant for case 3 is larger. It means this will represent the graph with more rate of decrease of current, that is, graph c.

Hence, case 1 will represent graph d.

In a similar way, case 2 will be represented by graph a and case 4 will be represented by graph b.

Hence, case 1 will represent graph d, case 2 will be represented by graph a, case 3 will be represented by graph c, case 4 will be represented by graph b.

Therefore, we can determine the nature of the graph by finding the current and time constant and then comparing them.

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