Chapter 18, Problem 11P

(a)

To determine

The resistance of a battery.

Answer to Problem 11P

  $\text{20}\text{Ω}$

Explanation of Solution

Given:

Voltage, $\text{V=12}\text{V}$ and current, $\text{I=0}\text{.60}\text{A}$ .

Formula used:

From the Ohm’s law,

The relation between resistance, voltage and current

  $\text{V=IR}$

Where $V=$ voltage, $I=$ current, $R=$ resistance.

After rearranging the expression: $\text{R=}\frac{\text{V}}{\text{I}}$

Calculation:

Substituting the values in the formula, $\text{R=}\frac{\text{V}}{\text{I}}$

  $\text{=}\frac{\text{12}\text{V}}{\text{0}\text{.60}\text{A}}$

  $\text{=}\text{20}\text{Ω}$

Conclusion:

The resistance is $\text{20}\text{Ω}$ .

(b)

To determine

The energy loss by battery in a minute.

Answer to Problem 11P

  $432J$

Explanation of Solution

Given:

Voltage, $\text{V=12}\text{V}$ and current, $\text{I=0}\text{.60}\text{A}$

Formula used:

The change in potential energy is $V\Delta Q$ .

Calculation:

By using the formula, $\Delta U=V\Delta Q=V\Delta t$

Calculation for the energy per minute

It is known that $\text{1 min}\text{= 60 s}$

So,

  $\Delta U=V\Delta Q$

  $=\left(0.60A\right)\left(60\mathrm{s}\right)\left(12volt\right)=432J$

Conclusion:

The energy loss by battery is $432J$ .