Chapter 18, Problem 35P

(a)

To determine

The resistance of the bulb and the power dissipated should be calculated.

Answer to Problem 35P

  $R\approx 6.7\Omega$

  $P\approx 1.4W$

Explanation of Solution

Given:

Current, $I=450mA=0.45Amp$

Voltage, $V=1.5Volt=2\times 1.5Volt=3.0Volt$

Formula used:

Expression to calculate voltage:

  $V=I\times R$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

General form of electrical power:

  $Electricpower=Voltagedifference\times current$

Calculation:

Rearrange the given formula, $V=I\times R$

So, a new expression is $R=\frac{V}{I}$

Substitute the given values,

  $R=\frac{3.0volt}{0.45Amp}=6.667\Omega \approx 6.7\Omega$

Now, calculate power by substituting the given values in formula

  $Electricpower=Voltagedifference\times current$

  $P=\left(0.45A\right)\left(3.0V\right)=1.35W\approx 1.4W$

Conclusion:

The resistance of the bulb is $R\approx 6.7\Omega$ and the power dissipated, $P\approx 1.4W$

(b)

To determine

Power would increase by a factor should be discussed.

Answer to Problem 35P

Power would increase by a factor of $4$ .

Explanation of Solution

Given:

Current, $I=450mA=0.45Amp$

Voltage, $V=1.5Volt=2\times 1.5Volt=3.0Volt$

Formula used:

Expression to calculate voltage:

  $V=I\times R$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

General form of electrical power:

  $Electricpower=Voltagedifference\times current$

Calculation:

Power would increase by a factor of $4$ . Using $4$ -cells would not be a good idea since the bulb will probably burn out.

Conclusion: The required factor is four.