Chapter 17, Problem 41P

Lightbulb A is marked “25.0 W 120. V,” and lightbulb B is marked “100. W 120. V.” These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120.-V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.

To determine

The resistance of light bulb A and light bulb B.

Answer to Problem 41P

The resistances of light bulb A and light bulb B is $576\Omega$ and $144\Omega$ respectively.

Explanation of Solution

Given Info: The power delivered to light bulb A when connected to $120\text{V}$ source is $25\text{W}$, The power delivered to bulb B is $100\text{W}$ when $120\text{V}$ is applied.

Explanation:

Formula to calculate resistance of bulb A is,

$R_{\text{A}}=\frac{V^2}{P_{\text{A}}}$

• $R_{\text{A}}$ is the resistance of bulb A,
• $V$ is the voltage of power supply,
• $P_{\text{A}}$ is the power delivered to bulb A,

Substitute $120\text{V}$ for $V$ and $25\text{W}$ for $R_{\text{A}}$ in the above equation to find $R_{\text{A}}$.

$\begin{array}{c}{R}_{\text{A}}=\frac{{\left(120\text{V}\right)}^2}{25\text{W}}\\ =576\Omega \end{array}$                                                                           (I)

Therefore, the resistance of the light bulb A is $576\Omega$.

Formula to calculate resistance of bulb B is,

$R_{\text{B}}=\frac{V^2}{P_{\text{B}}}$

• $R_{\text{B}}$ is the resistance of bulb B,
• $V$ is the voltage of power supply,
• $P_{\text{B}}$ is the power delivered to bulb B,

Substitute $120\text{V}$ for $V$ and $100\text{W}$ for $P_{\text{B}}$ in the above equation to find $R_{\text{B}}$,

$\begin{array}{c}{R}_B=\frac{{\left(120\text{V}\right)}^2}{100\text{W}}\\ =144\Omega \end{array}$ (II)

Therefore, the resistance of the light bulb B is $144\Omega$

Conclusion:

Therefore, from equation (I) resistance of light bulb A is $576\Omega$ and from equation (II) light bulb B has resistance $144\Omega$.

To determine

The time taken by $1.00\text{C}$ charge to pass into light bulb A.

Answer to Problem 41P

The time taken by $1.00\text{C}$ charge to pass into light bulb A is $4.80\text{s}$.

Explanation of Solution

Given Info: The power delivered to light bulb A when connected to $120\text{V}$ source is $25\text{W}$. Power delivered to bulb B is $100\text{W}$ on applying $120\text{V}$. The charge pass into light bulb A is $1.00\text{C}$.

Explanation:

Formula to calculate the time taken is,

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

• $\Delta t$ is the time taken by charge to pass into light bulb A,
• $Q$ is the amount of charge,
• $I$ is the current produced,

Equation for current is given by,

$I=\frac{V}{R}$

• $V$ is the voltage of power supply,
• $R$ is the resistance of light bulb A,

Substitute the above equation in the previous equation and rewrite $\Delta t$,

$\Delta t=\frac{QR}{V}$

Substitute $1.00\text{C}$ for $Q$ , $576\Omega$ for $R$ and $120\text{V}$ for $V$ in the above equation to find $\Delta t$.

$\begin{array}{c}\Delta t=\frac{\left(1.00\text{C}\right)\left(576\Omega \right)}{120\text{V}}\\ =4.80\text{s}\end{array}$

Conclusion:

Therefore, the time taken by $1.00\text{C}$ charge to pass into light bulb A is $4.80\text{s}$

To determine

Whether the charge is different upon its exit versus entry into the light bulb.

Answer to Problem 41P

Answer: No. The charge entered into bulb is same as that of charge comes out from the bulb.

Explanation of Solution

By law of conservation of charges, total number of charges remains the same.

Charge is a conserved quantity. It can be neither created nor destroyed. So the number of charges entered into light bulb must be same as that of charges exit from the bulb.

Conclusion: No. The charge entered into bulb is same as that of charge comes out from the bulb.

To determine

The time interval for $1.00\text{J}$ pass into light bulb A.

Answer to Problem 41P

The time interval for $1.00\text{J}$ pass into light bulb A is $0.040\text{s}$.

Explanation of Solution

Given Info: The power delivered to light bulb A when connected to $120\text{V}$ source is $25\text{W}$ . Power delivered to bulb B is $100\text{W}$ on applying $120\text{V}$. The energy passes into light bulb A is $1.00\text{J}$.

Explanation:

Formula to calculate the time interval for energy to pass is,

$\Delta t=\frac{E}{P_{\text{A}}}$

• $E$ is the energy passes into light bulb A,

Substitute $1.00\text{J}$ for $E$ and $25\text{W}$ in the above equation to find $\Delta t$,

$\begin{array}{c}\Delta t=\frac{1.00\text{J}}{25\text{W}}\\ =0.040\text{s}\end{array}$

Conclusion: Therefore, the time interval for $1.00\text{J}$ pass into light bulb A is $0.040\text{s}$.

To determine

Mechanisms with which energy enters and leaves the light bulb.

Answer to Problem 41P

Energy enters in form of electric current and exit as light and heat energy forms.

Explanation of Solution

The form of energy which exit the bulb is different from the form of energy which enters the light bulb.

The energy enters the light bulb is in the form of electric current while the energy comes out is in the form of light and heat energy.

Conclusion: Energy enters in form of electric current and exit as light and heat energy forms.

To determine

The cost of running light bulb A

Answer to Problem 41P

The cost of running light bulb A $\$1.98$

Explanation of Solution

Given Info: The power delivered to light bulb A when connected to $120\text{V}$ source is $25\text{W}$. Power delivered to bulb B is $100\text{W}$ on applying $120\text{V}$. The light bulb A is used for 30 days. The cost of $1\text{kWh}$ is $\$0.110$

Explanation:

Formula to calculate the total energy used is,

$E=P_{\text{A}}\Delta t$

• $E$ is the total energy consumed by light bulb A,

Substitute $25\text{W}$ for $P_{\text{A}}$ and $30\text{days}$ for $\Delta t$ in the above equation to find $E$,

$\begin{array}{c}E=\left(25\text{W}\right)\left(\frac{{\text{10}}^{\text{-3}}\text{kW}}{\text{1}\text{W}}\right)\left(\text{30}\text{days}\right)\left(\frac{\text{24}\text{hours}}{\text{1}\text{day}}\right)\\ =18.0\text{kWh}\end{array}$

Conclusion:

The total energy consumed by light bulb A in 30 days is $18.0\text{kWh}$.

The total cost of running the light bulb A for 30 days such that cost of $1\text{kWh}$ is $\$0.110$ is given by,

$\begin{array}{c}\text{Cost}=\left(18.0\text{kWh}\right)\left(\frac{\$0.110}{\text{kWh}}\right)\\ =\$1.98\end{array}$

Therefore, the total cost for running bulb is $\$1.98$