Question

After the switch in Fig. 27-15 is closed on point a, there is current ithrough resistance R. Figure 27-23 gives that current for four sets of values of Rand 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?

Short answer

Nature of graph for the given conditions is as follows:

1-(d)

2 -(a)

3-(c)

4- (b)

Step-by-step solution

Step 1: Given

$$\begin{gathered} {\mathrm{R}}_0\mathrm{and}{\mathrm{C}}_0 \\ 2{\mathrm{R}}_0\mathrm{and}{\mathrm{C}}_0 \\ {\mathrm{R}}_0\mathrm{and}2{\mathrm{C}}_0 \\ 2{\mathrm{R}}_0\mathrm{and}2{\mathrm{C}}_0 \end{gathered}$$

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:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ \mathrm{V}=\mathrm{IR} \\ \mathrm{i}=-\frac{\mathrm{q}}{\mathrm{RC}}*{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{RC}}} \end{gathered}$$

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

Determining the nature of graph for the given conditions

$\mathrm{Case}1){\mathrm{R}}_0\mathrm{and}{\mathrm{C}}_0$

First findthecharge as follows:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ \mathrm{q}={\mathrm{C}}_0{\mathrm{V}}_0 \end{gathered}$$

Current is given as follows:

$$\begin{gathered} {\mathrm{I}}_0=\frac{{\mathrm{q}}_0}{{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{C}}_0{\mathrm{V}}_0}{{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{{\mathrm{R}}_0} \end{gathered}$$

The time constant for this case will be ,

$\mathrm{t}={\mathrm{R}}_0{\mathrm{C}}_0$

$\mathrm{Case}2)2{\mathrm{R}}_0\mathrm{and}{\mathrm{C}}_0$

First findthecharge as follows:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ {\mathrm{q}}_0={\mathrm{C}}_0{\mathrm{V}}_0 \end{gathered}$$

Current is given as follows:

$$\begin{gathered} {\mathrm{I}}_0=\frac{{\mathrm{q}}_0}{2{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{C}}_0{\mathrm{V}}_0}{2{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{2{\mathrm{R}}_0} \end{gathered}$$

The time constant is,

$\mathrm{t}=\mathrm{RC}=2{\mathrm{R}}_0{\mathrm{C}}_0$

$\mathrm{Case}3){\mathrm{R}}_0\mathrm{and}2{\mathrm{C}}_0$

First findthecharge as follows:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ {\mathrm{q}}_0=2{\mathrm{C}}_0{\mathrm{V}}_0 \end{gathered}$$

Current is given as follows:

$$\begin{gathered} {\mathrm{I}}_0=\frac{{\mathrm{q}}_0}{2{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{2{\mathrm{C}}_0{\mathrm{V}}_0}{2{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{{\mathrm{R}}_0} \end{gathered}$$

The time constant for this case is,

$\mathrm{t}=\mathrm{RC}=2{\mathrm{R}}_0{\mathrm{C}}_0$

$\mathrm{Case}4)2{\mathrm{R}}_0\mathrm{and}2{\mathrm{C}}_0$

First findthecharge as follows:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ \mathrm{q}=2{\mathrm{C}}_0{\mathrm{V}}_0 \end{gathered}$$

Current is given as follows:

$$\begin{gathered} {\mathrm{I}}_0=\frac{{\mathrm{q}}_0}{4{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{2{\mathrm{C}}_0{\mathrm{V}}_0}{4{\mathrm{R}}_0{\mathrm{C}}_0} \\ {\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{2{\mathrm{R}}_0} \end{gathered}$$

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,thevalue of time constant for case 3 is larger. It means this will representthegraph 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.