Chapter 27, Problem 3OQ

To determine

The ranking of the given actions in terms of the change in current from greatest increase to the greatest decrease.

Answer to Problem 3OQ

The ranking of current in decreasing order is given by $\left(\text{c}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{e}\right)>\left(\text{d}\right)$.

Explanation of Solution

Write the expression for the Ohm’s Law.

    $V=IR$                                                                                                               (I)

Here, $R$ is the resistance, $V$ is the voltage and $I$ is the current.

Write the expression for the power in terms of voltage.

    $P=\frac{V^2}{R}$                                                                                                            (II)

Here, $P$ is the power.

Write the expression for the resistance.

    $R=\rho \frac{l}{A}$                                                                                                         (III)

Here, $R$ is the resistance, $\rho$ is the resistivity, $l$ is the length and $A$ is the cross sectional area of the wire.

Write the expression for the current in terms of potential difference.

    $I=\frac{V_{\text{A}}-V_{\text{B}}}{R}$                                                                                                   (IV)

Here, $V_{\text{A}}$ is the potential at point A and $V_{\text{B}}$ is the potential at point B.

Write the expression for the resistance of the wire in terms of temperature.

    $R=R_0\left(1+\alpha T\right)$                                                                                               (V)

Here, $R$ is the resistance at temperature $T$ and $\alpha$ is the coefficient of temperature.

Case (a):

Substitute $50\text{V}$ for $V_{\text{A}}$ and $0\text{V}$ for $V_{\text{B}}$ in equation (IV) to find $I_1$.

    $\begin{array}{c}{I}_1=\frac{50\text{V}-0\text{V}}{R}\\ =\frac{50\text{V}}{R}\end{array}$

Substitute $150\text{V}$ for $V_{\text{A}}$ and $0\text{V}$ for $V_{\text{A}}$ in the equation (IV) to find $I_2$.

    $\begin{array}{c}{I}_2=\frac{150\text{V}-0\text{V}}{R}\\ =\frac{150\text{V}}{R}\\ =\frac{3\times 50\text{V}}{R}\end{array}$

Substitute $\frac{50\text{V}}{R}$ for $I_1$ in the above equation to find $I_1$.

    $I_2=3I_1$

Therefore, the current is $I_2=3I_1$.

Case (b):

Rearrange equation (II) to find voltage.

    $V=\sqrt{PR}$

Write the expression for the ratio of the voltages at power $P_1$ and $P_2$.

    $\frac{V_1}{V_2}=\sqrt{\frac{P_1}{P_2}}$

Substitute $3P_1$ for $P_2$ in the above equation.

    $\begin{array}{c}\frac{V_1}{V_2}=\sqrt{\frac{P_1}{3P_1}}\\ \frac{V_1}{V_2}=\sqrt{\frac{1}{3}}\\ V_2=\sqrt{3}{V}_1\end{array}$

Substitute $\frac{I_2}{R}$ for $V_2$ and $\frac{I_1}{R}$ for $V_1$ in the above equation to find $I_2$.

    $I_2=I_1\sqrt{3}$

Therefore, the value of current for triple power is $I_2=I_1\sqrt{3}$.

Case (c):

Substitute equation (III) in equation (I) to find $V_1$.

    $V_1=\frac{\rho I_{1}l}{A_1}$

Substitute $\pi r_1^2$ for $A_1$ in the above equation.

    $V_1=\frac{\rho I_{1}l}{\pi r_1^2}$

Write the expression for voltage $V_2$.

    $V_2=\frac{\rho I_{2}l}{\pi r_2^2}$

Write the expression for the ratio of $V_1$ and $V_2$.

    $\begin{array}{c}\frac{V_1}{V_2}=\frac{\left(\frac{\rho I_{1}l}{\pi r_1^2}\right)}{\left(\frac{\rho I_{2}l}{\pi r_2^2}\right)}\\ =\frac{r_2^2{I}_2}{r_1^2{I}_1}\end{array}$

Substitute $2r_1$ for $r_2$ in the above expression.

    $\frac{V_1}{V_2}=\frac{4I_2}{I_1}$

Therefore, the current is $I_2=4I_1$.

Case (d):

Substitute equation (III) in equation (I) to find $V_1$.

    $V_1=\frac{\rho I_{1}l}{A_1}$

Write the expression for voltage $V_2$.

    $V_2=\frac{\rho I_{2}l}{A_2}$

Write the expression for the ratio of voltage $V_1$ and $V_2$.

    $\begin{array}{c}\frac{V_1}{V_2}=\frac{\left(\frac{\rho I_{1}l}{A_1}\right)}{\left(\frac{\rho I_{2}l}{A_2}\right)}\\ =\frac{I_1{l}_1}{I_2{l}_2}\end{array}$

Substitute $2l_1$ for $l_2$ in the above equation.

    $\begin{array}{c}\frac{V_1}{V_2}=\frac{I_1{l}_1}{I_2\cdot 2l_1}\\ =\frac{I_1}{2I_2}\end{array}$

Therefore, then current is $I_2=\frac{1}{2}{I}_1$.

Case (e):

From equation (V), it is clear that resistance is proportional to temperature $T$.

Hence, when temperature is doubled, the resistance is also doubled.

The ratio of voltages from equation (I) is given by:

    $\frac{V_1}{V_2}=\frac{{{\rm I}}_1{R}_1}{I_2{R}_2}$

Substitute $2R_1$ for $R_2$ in the above equation, we get,

    $\begin{array}{c}\frac{V_1}{V_2}=\frac{{{\rm I}}_1{R}_1}{I_2\cdot 2R_1}\\ =\frac{I_1}{2I_2}\end{array}$

Therefore, the current is $I_2=\frac{1}{2}{I}_1$.

Conclusion:

Therefore, the ranking of current in the decreasing order is given by $\left(\text{c}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{e}\right)>\left(\text{d}\right)$.