Chapter 19, Problem 66GP

To determine

The voltage drop across each connecting wire and voltage applied to apparatus.

Answer to Problem 66GP

The voltage drop across connecting wire is $1.82\text{V}$ and the voltage applied to apparatus is $121.85\text{V}$ .

Explanation of Solution

Given info:

The voltage of the source is, $V=120\text{V}$ .

The length of the wire is, $l=95\text{m}$ .

The resistance per unit length of the wire is, $r=0.0065\Omega /\text{m}$ .

The current drawn by the apparatus is, $I=3\text{A}$ .

Formula Used:

The expression to calculate the resistance of the connecting wire is,

  $R_w=l\cdot r$    ........(1)

The expression to calculate the resistance of the apparatus is,

  $R=\frac{V}{I}$    ........(2)

The expression to calculate the voltage drop across connecting wire is,

  $V_w=\left(\frac{R_w}{R+R_w}\right)V$    ........(3)

The expression to calculate the voltage applied to apparatus is,

  $V_a=I\left(R+r\right)$    ........(4)

Calculation:

Substitute all the values in the above expression (1).

  $\begin{array}{c}{R}_w=\left(95\text{m}\right)\left(0.0065\Omega /\text{m}\right)\\ =0.6175\Omega \end{array}$

Substitute all the values in the above expression (2).

  $\begin{array}{c}R=\frac{120\text{V}}{3\text{A}}\\ =40\Omega \end{array}$

Substitute all the values in the above expression (3).

  $\begin{array}{c}{V}_w=\left(\frac{0.6175\Omega }{40\Omega +0.6175\Omega }\right)\left(120\text{V}\right)\\ =1.82\text{V}\end{array}$

Substitute all the values in the above expression (4).

  $\begin{array}{c}{V}_a=\left(3\text{A}\right)\left(40\Omega +0.6175\Omega \right)\\ =121.85\text{V}\end{array}$

Conclusion:

Thus, the voltage drop across connecting wire is $1.82\text{V}$ and the voltage applied to apparatus is $121.85\text{V}$ .