Chapter 26, Problem 34AP

Lightbulb A is marked “25 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 each light bulb.

Answer to Problem 34AP

The resistance of lightbulb A is $576\text{Ω}$ and resistance of lightbulb B is $144\text{Ω}$.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$.

Explanation:

Formula to calculate the resistance of lightbulb A.

$\begin{array}{c}{P}_A=\frac{V^2}{R_A}\\ R_A=\frac{V^2}{P_A}\end{array}$ (1)

Here,

$P_A$ is the power of light bulb A.

$R_A$ is the resistance of lightbulb A.

$V$ is the voltage across light bulb A.

Substitute $120\text{V}$ for $V$, $25\text{W}$ for $P_A$ in equation (1) to find $R_A$,

$\begin{array}{c}{R}_A=\frac{{\left(120\text{V}\right)}^2}{25\text{W}}\\ =576\text{Ω}\end{array}$

Thus, the resistance of lightbulb A is $576\text{Ω}$.

Formula to calculate the resistance of lightbulb B.

$\begin{array}{c}{P}_B=\frac{V^2}{R_B}\\ R_B=\frac{V^2}{P_B}\end{array}$ (2)

Here,

$P_B$ is the power of light bulb B.

$R_B$ is the resistance of lightbulb B.

$V$ is the voltage across light bulb B.

Substitute $120\text{V}$ for $V$, $100\text{W}$ for $P_B$ in equation (2) to find $R_B$,

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

Thus, the resistance of lightbulb B is $144\text{Ω}$.

Conclusion:

Therefore, the resistance of lightbulb A is $576\text{Ω}$ and resistance of lightbulb B is $144\text{Ω}$.

To determine

The time interval through which $1.00\text{C}$ pass into light bulb A.

Answer to Problem 34AP

The time interval through which $1.00\text{C}$ pass into light bulb A is $4.808\text{s}$.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$, charge across that passes through light bulb A is $1.00\text{C}$.

Explanation:

Formula to calculate the current flowing in th light bulb A.

$I_A=\frac{V}{R_A}$ (3)

Here,

$I_A$ is the current flowing in th light bulb A.

Substitute $120\text{V}$ for $V$, $576\text{Ω}$ for $R_A$ in equation (4) to find $I_A$,

$\begin{array}{c}{I}_A=\frac{120\text{V}}{576\text{Ω}}\\ =0.208\text{A}\end{array}$

Thus, the current flowing in th light bulb A is $0.208\text{A}$.

Formula to calculate the time interval through which $1.00\text{C}$ pass into light bulb A.

$\begin{array}{c}{I}_A=\frac{Q_{1.00\text{C}}}{t_{1.00\text{C}}}\\ t_{1.00\text{C}}=\frac{Q_{1.00\text{C}}}{I_A}\end{array}$ (4)

Here,

$t_{1.00\text{C}}$ is the time interval through which $1.00\text{C}$ pass into light bulb A.

$Q_{1.00\text{C}}$ is the charge across that passes through light bulb A.

Substitute $1.00\text{C}$ for $Q_{1.00\text{C}}$, $0.208\text{A}$ for $I_A$ in equation (4) to find $t_{1.00\text{C}}$,

$\begin{array}{c}{t}_{1.00\text{C}}=\frac{1.00\text{C}}{0.208\text{A}}\\ =4.8076\text{s}\\ \approx 4.808\text{s}\end{array}$

Thus, the time interval through which $1.00\text{C}$ pass into light bulb A is $4.807\text{s}$.

Conclusion:

Therefore, the time interval through which $1.00\text{C}$ pass into light bulb A is $4.807\text{s}$.

To determine

The reason that this charge is different upon its exit versus its entry into the light bulb or not.

Answer to Problem 34AP

This charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$, charge across that passes through light bulb A is $1.00\text{C}$.

Explanation:

No, the existing charge is the same amount as the entering charge into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Thus, this charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Conclusion:

Therefore, this charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

To determine

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

Answer to Problem 34AP

The time interval through which $1.00\text{J}$ pass into light bulb A is $0.04\text{s}$.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$, energy that passes through light bulb A is $1.00\text{J}$.

Explanation:

Formula to calculate the time interval through which $1.00\text{J}$ pass into light bulb A.

$\begin{array}{c}{E}_A=P_A\times t_{1.00\text{J}}\\ t_{1.00\text{J}}=\frac{E_A}{P_A}\end{array}$ (5)

Here,

$t_{1.00\text{J}}$ is the time interval through which $1.00\text{J}$ pass into light bulb A.

$E_A$ is the energy that passes through light bulb A

Substitute $1.00\text{J}$ for $E_A$, $25\text{W}$ for $P_A$ in equation (5) to find $t_{1.00\text{J}}$,

$\begin{array}{c}{t}_{1.00\text{J}}=\frac{1.00\text{J}}{25\text{W}}\\ =0.04\text{s}\end{array}$

Thus, the time interval through which $1.00\text{J}$ pass into light bulb A is $0.04\text{s}$.

Conclusion:

Therefore, the time interval through which $1.00\text{J}$ pass into light bulb A is $0.04\text{s}$.

To determine

The mechanism through which this energy enter and exit the light bulb.

Answer to Problem 34AP

The mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$, energy that passes through light bulb A is $1.00\text{J}$.

Explanation:

In this mechanism, the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Thus, the mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Conclusion:

Therefore, the mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

To determine

The cost of running light bulb A continuously for $30.0\text{days}$.

Answer to Problem 34AP

The cost of running light bulb A continuously for $30.0\text{days}$ is $\$1.98$.

Explanation of Solution

Given information: Power of light bulb A is $25\text{W}$, and voltage across light bulb A is $120\text{V}$, power of light bulb B is $100\text{W}$, and voltage across light bulb B is $120\text{V}$, constant voltage source is $120\text{V}$, energy that passes through light bulb A is $1.00\text{J}$, running time of light bulb A is $30.0\text{days}$, selling cost at which electric company sells product is $\$0.110/\text{kWh}$.

Explanation:

Write the expression for the energy for light bulb A works continuously for $30.0\text{days}$.

$E_{30.0\text{days}}=P\times t$ (6)

Here,

$E_{30.0\text{days}}$ is the energy for light bulb A works continuously for $30.0\text{days}$.

$P$ is the power of light bulb A.

$t$ is the running time of light bulb A.

Substitute $25\text{W}$ for $P$, $30.0\text{days}$ for $t$ in equation (6) to find $E_{30.0\text{days}}$,

$\begin{array}{c}{E}_{30.0\text{days}}=25\text{W}\times \left(30.0\text{days}\times \frac{24\text{h}}{1\text{day}}\right)\\ =18000\text{Wh}\\ =18\text{kWh}\end{array}$

Thus, the energy for light bulb A works continuously for $30.0\text{days}$ is $18\text{kWh}$.

Formula to calculate the cost of running light bulb A continuously for $30.0\text{days}$.

$C_R=E_{30.0\text{days}}\times C_S$ (7)

Here,

$C_R$ is the cost of running light bulb A continuously for $30.0\text{days}$.

$C_S$ is the selling cost at which electric company sells product.

Substitute $18\text{kWh}$ for $E_{30.0\text{days}}$, $\$0.110/\text{kWh}$ for $C_S$ in equation (7) to find $C_R$,

$\begin{array}{c}{C}_R=18\text{kWh}\times \$0.110/\text{kWh}\\ =\$1.98\end{array}$

Thus, the cost of running light bulb A continuously for $30.0\text{days}$ is $\$1.98$.

Conclusion:

Therefore, the cost of running light bulb A continuously for $30.0\text{days}$ is $\$1.98$.