Chapter 1.3, Problem 33E

a.

To determine

To find: the effective resistance of battery.

Answer to Problem 33E

  $0.4\text{ ohms}$

Explanation of Solution

Given information:

The given variables are as follows:

Voltage (in volts) = $V$

Current drawn (in amperes) = $i$

Effective resistance (in ohms) = $R$

The slope of function $V\left(i\right)=m$ , and $R=-m$ .

  $\begin{array}{l}\left(i_1,V_1\right)=\left(1.0,12.0\right)\\ \left(i_2,V_2\right)=\left(10.0,8.4\right)\end{array}$

Calculation:

Put the values of $\left(i_1,V_1\right)\text{ and }\left(i_2,V_2\right)$ in the slope formula and evaluate the value of m as follows:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{8.4-12.0}{10.0-1.0}\\ =\frac{-3.6}{9}\\ =-0.4\end{array}$

Put the value of m in the following equation and evaluate R as follows:

  $\begin{array}{c}R=-m\\ =-\left(-0.4\right)\\ =0.4\end{array}$

Thus, the effective resistance is 0.4 ohms.

b.

To determine

To find: the voltage produced by the battery at given current supply.

Answer to Problem 33E

2.4 Volts

Explanation of Solution

Given information:

The given variables are as follows:

Voltage (in volts) = $V$

Current drawn (in amperes) = $i$

Effective resistance (in ohms) = $R$

The slope of function $V\left(i\right)=m$ , and $R=-m$ .

  $\begin{array}{l}\left(i_1,V_1\right)=\left(1.0,12.0\right)\\ \left(i_2,V_2\right)=\left(10.0,8.4\right)\end{array}$

  $\begin{array}{l}m=-0.4\\ R=0.4\text{ ohms}\\ i=25.0\text{ ampere}\end{array}$

Calculation:

Since the function of the line represented by $V\left(i\right)$ . The slope intercept equation for this function will be as follows:

  $\begin{array}{c}y=mx+b\\ V\left(i\right)=-0.4i+b\mathrm{...}\left(1\right)\end{array}$

Put the first point coordinates $\left(i_1,V_1\right)$ in equation (1) and evaluate b as follows:

  $\begin{array}{c}12.0=\left(-0.4\right)\left(1.0\right)+b\\ 12.0=-0.4+b\\ b=12.0+0.4\\ b=12.4\end{array}$

Now, put the value of b in equation (1) to evaluate the slope intercept equation $V\left(i\right)$ as follows:

  $V\left(i\right)=-0.4i+12.4\mathrm{...}\left(2\right)$

Now, put $i=25.0\text{ ampere}$ in equation (2) to evaluate the voltage of the battery as shown below.

  $\begin{array}{c}V\left(i\right)=-0.4\left(25.0\right)+12.4\\ =-10+12.4\\ =2.4\end{array}$

Thus, the voltage produced by battery at given current supply is 2.4 volts.