Chapter 32, Problem 1P

(a) What is the resistance of a lightbulb that uses an average power of 75.0 W when connected to a 60.0-Hz power source having a maximum voltage of 170 V? (b) What If? What is the resistance of a 100-W lightbulb?

To determine

The resistance of a light bulb.

Answer to Problem 1P

The resistance of a light bulb is $193\text{Ω}$.

Explanation of Solution

Given Information: Average power of the light bulb is $75.0\text{W}$, frequency of the power source is $60.0\text{Hz}$, maximum voltage across the power source is $170\text{V}$.

Formula to calculate the root mean square voltage across the power source is,

$V_{\text{rms}}=\frac{V_{\text{max}}}{\sqrt{2}}=0.707V_{\text{max}}$ (1)

Here,

$V_{\text{rms}}$ is the root mean square voltage across the power source.

$V_{\text{max}}$ is the maximum voltage across the power source.

Substitute $170\text{V}$ for $V_{\text{max}}$ in equation (1) to find $V_{\text{rms}}$,

$\begin{array}{c}{V}_{\text{rms}}=\frac{170\text{V}}{\sqrt{2}}\\ =120.21\text{V}\end{array}$

Thus, the root mean square voltage across the power source is $120.21\text{V}$.

Formula to calculate the root mean square current across the power source is,

$I_{\text{rms}}=\frac{P}{V_{\text{rms}}}$ (2)

Here,

$I_{\text{rms}}$ is the root mean square current across the power source.

$P$ is the average power of the light bulb.

Substitute $120.21\text{V}$ for $V_{\text{rms}}$, $75.0\text{W}$ for $P$ in equation (2) to find $I_{\text{rms}}$,

$\begin{array}{c}{I}_{\text{rms}}=\frac{75.0\text{W}}{120.21\text{V}}\\ =0.623\text{A}\end{array}$

Thus, the value of the root mean square current across the power source is $0.623\text{A}$.

Formula to calculate the resistance of a light bulb is,

$R=\frac{V_{\text{rms}}}{I_{\text{rms}}}$ (3)

Here,

$R$ is the resistance of a light bulb.

Substitute $120.21\text{V}$ for $V_{\text{rms}}$, $0.623\text{A}$ for $I_{\text{rms}}$ in equation (3) to find $R$,

$\begin{array}{c}R=\frac{120.21\text{V}}{0.623\text{A}}\\ =192.67\text{Ω}\\ \approx 193\text{Ω}\end{array}$

Thus, the resistance of a light bulb is $193\text{Ω}$.

Conclusion:

Therefore, the resistance of a light bulb is $193\text{Ω}$.

To determine

The resistance of a $100\text{W}$ light bulb.

Answer to Problem 1P

The resistance of a $100\text{W}$ light bulb is $144\text{Ω}$.

Explanation of Solution

Given Information: Average power of the light bulb is $100\text{W}$, frequency of the power source is $60.0\text{Hz}$, maximum voltage across the power source is $170\text{V}$.

Formula to calculate the root mean square current across the power source is,

${I^{\prime }}_{\text{rms}}=\frac{P^{\prime }}{V_{\text{rms}}}$ (4)

Here,

${I^{\prime }}_{\text{rms}}$ is the root mean square current across the power source.

$P^{\prime }$ is the average power of the $100\text{W}$ light bulb.

Substitute $120.21\text{V}$ for $V_{\text{rms}}$, $100\text{W}$ for $P^{\prime }$ in equation (2) to find $I_{\text{rms}}$,

$\begin{array}{c}{I^{\prime }}_{\text{rms}}=\frac{100\text{W}}{120.21\text{V}}\\ =0.8318\text{A}\\ \approx 0.832\text{A}\end{array}$

Thus, the value of the root mean square current across the power source is $0.832\text{A}$.

Formula to calculate the resistance of a light bulb is,

$R=\frac{V_{\text{rms}}}{{I^{\prime }}_{\text{rms}}}$ (5)

Substitute $120.21\text{V}$ for $V_{\text{rms}}$, $0.832\text{A}$ for ${I^{\prime }}_{\text{rms}}$ in equation (3) to find $R$,

$\begin{array}{c}R=\frac{120.21\text{V}}{0.832\text{A}}\\ =144.4\text{Ω}\\ \approx 144\text{Ω}\end{array}$

Thus, the resistance of a $100\text{W}$ light bulb is $144\text{Ω}$.

Conclusion:

Therefore, the resistance of a $100\text{W}$ light bulb is $144\text{Ω}$.