Chapter 18, Problem 79GP

(a)

To determine

The current though each bulb.

Answer to Problem 79GP

The current in a lightbulb A is $0.33A$

The current in a lightbulb B is $3.33A$

Explanation of Solution

Given:

Power lightbulb A, $P_A=40W$

Voltage of lightbulb A, $V_A=120V$

Power lightbulb B, $P_B=40W$

Voltage of lightbulb B, $V_B=12V$

Formula used:

  $P=VI$

Where, $P$ is power, $V$ is voltage and $I$ is current.

Calculation:

Rearrange the formula of power, $P=VI$ . So, $I=\frac{P}{V}$

According to the question, current in a lightbulb A:

  $I_A=\frac{P_A}{V_A}=\frac{40}{120}=0.33A$

According to the question, current in a lightbulb B:

  $I_B=\frac{P_B}{V_B}=\frac{40}{12}=3.33A$

Conclusion: The current in a lightbulb A is $0.33A$ and the current in a lightbulb B is $3.33A$

(b)

To determine

The resistance though each bulb.

Answer to Problem 79GP

The resistance in a lightbulb A is $360\Omega$

The resistance in a lightbulb B is $3.6\Omega$

Explanation of Solution

Given:

Power lightbulb A, $P_A=40W$

Voltage of lightbulb A, $V_A=120V$

Power lightbulb B, $P_B=40W$

Voltage of lightbulb B, $V_B=12V$

Formula used:

Power

  $P=VI$

Where, $V$ is voltage and $I$ is current.

Also, $P=\frac{V^2}{R}$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

Calculation:

Rearrange the formula, $P=\frac{V^2}{R}$ . So, $R=\frac{V^2}{P}$

According to the question, resistance in a lightbulb A:

  $R_A=\frac{V_A^2}{P_A}=\frac{{120}^2}{40}=360\Omega$

According to the question, resistance in a lightbulb B:

  $R_B=\frac{V_B^2}{P_B}=\frac{{12}^2}{40}=3.6\Omega$

Conclusion: The resistance in a lightbulb A is $360\Omega$ and the resistance in a lightbulb B is $3.6\Omega$

(c)

To determine

The amount of chargethat passes in one hour.

Answer to Problem 79GP

The charge in the lightbulb A is $1.2\times {10}^{3}C$

The charge in a lightbulb B is $1.2\times {10}^{4}C$ .

Explanation of Solution

Given:

Power lightbulb A, $P_A=40W$

Voltage of lightbulb A, $V_A=120V$

Power lightbulb B, $P_B=40W$

Voltage of lightbulb B, $V_B=12V$

Current in lightbulb A, $I_A=0.33A$

Current in lightbulb B, $I_B=3.33A$

Temperature change, $\Delta t=1.0h=3600s$

Formula used:

Current: $I=\frac{Q}{\Delta t}$

Where,

Q is the amount of charge and $\Delta t$ is the time interval.

Calculation:

Rearrange the formula, $I=\frac{Q}{\Delta t}$ . So, $Q=I\Delta t$

According to the question, amount of charge that passes in the lightbulb A:

  $Q_A=I_A\Delta t$

Substitute the given values:

  $Q_A=0.33\times 3600=1.2\times {10}^{3}C$

According to the question, amount of charge that passesin the lightbulb B:

  $Q_B=I_B\Delta t$

Substitute the given values:

  $Q_B=3.33\times 3600=1.2\times {10}^{4}C$

Conclusion: The charge that passesin lightbulb A is $1.2\times {10}^{3}C$ and the charge that passesin lightbulb B is $1.2\times {10}^{4}C$ .

(d)

To determine

The energy each bulb uses in one hour should be determined.

Answer to Problem 79GP

The answer is $1.44\times {10}^{5}J$ .

Explanation of Solution

Given:

Power lightbulb A, $P_A=40W$

Voltage of lightbulb A, $V_A=120V$

Power lightbulb B, $P_B=40W$

Voltage of lightbulb B, $V_B=12V$

Current in lightbulb A, $I_A=0.33A$

Current in lightbulb B, $I_B=3.33A$

Temperature change, $\Delta t=1.0h=3600s$

Formula used:

The power consumed by the light: $P=\frac{\Delta E}{\Delta t}$

Where,

  $\Delta E=consumedenergy$

  $\Delta t=temperaturechange$

  $P=Power$

Calculation:

Rearrange the formula, $P=\frac{\Delta E}{\Delta t}$ . So, $\Delta E=P\Delta t$

According to the question, two light bulbs are having the same power. So,

  $\Delta E_A=\Delta E_B=P\Delta t$

Substitute the given values:

  $\Delta E_A=\Delta E_B=40\times 3600s$

  $\Delta E_A=\Delta E_B=1.44\times {10}^{5}J$

Conclusion: The energy is $1.44\times {10}^{5}J$ .

(e)

To determine

The bulb that requires larger diameter wires to connect its power source and the bulb.

Answer to Problem 79GP

The diameter of wire B must be as larger because it has much resistance.

Explanation of Solution

Given:

Power lightbulb A, $P_A=40W$

Voltage of lightbulb A, $V_A=120V$

Power lightbulb B, $P_B=40W$

Voltage of lightbulb B, $V_B=12V$

Current in lightbulb A, $I_A=0.33A$

Current in lightbulb B, $I_B=3.33A$

Temperature change, $\Delta t=1.0h=3600s$

The resistance in a lightbulb A is $360\Omega$

The resistance in a lightbulb B is $3.6\Omega$

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

  $\pi$ is constant

  $r$ is a radius.

  $d$ is a diameter.

Calculation:

According to the given formula, $R=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}$ , the resistance increases when the diameter decreases. So, the wire that carries more current must have the largest diameter. It is known that increasing the resistance will increase the resistance of the heat. We need to make the current moves safely, so we must increase the diameter of the wire.

Since wire B has much resistance, its diameter must be as larger as possible.

Conclusion: The wire B has much resistance.