Chapter 23, Problem 67A

(a)

To determine

The resistance of the television.

Answer to Problem 67A

The resistance of the television is $52\Omega$ .

Explanation of Solution

Given:

The power of the television is $P=275\text{ W}$ .

The voltage is $V=120\text{ V}$ .

Formula used:

The expression for the power is,

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

Here, $V$ is the voltage and $R$ is the resistance.

Calculation:

The resistance of the television is,

$\begin{array}{l}P=\frac{V^2}{R}\\ R=\frac{{\left(120\text{ V}\right)}^2}{275\text{ W}}\\ R=52\Omega \end{array}$

Conclusion:

Thus, the resistance of the television is $52\Omega$ .

(b)

To determine

The voltage drop across the television.

Answer to Problem 67A

The voltage drop across the television is $115\text{ V}$ .

Explanation of Solution

Given:

The power of the television is $P=275\text{ W}$ .

The voltage is $V=120\text{ V}$ .

The resistance of the wire is $R_{\text{wire}}=2.5\text{Ω}$ .

Formula used:

The expression for the voltage drop is,

$V_{\text{B}}=\frac{V{R}_{\text{B}}}{R_{\text{A}}+R_{\text{B}}}$

Here, $V$ is the voltage, $R_{\text{A}}$ is the resistance and $R_B$ is the resistance.

Calculation:

Refer part (a).

The resistance of the television is $R_{\text{TV}}=52\Omega$ .

The voltage drop across the television is,

$\begin{array}{l}{V}_{\text{TV}}=\frac{V{R}_{\text{TV}}}{R_{\text{wire}}+R_{\text{TV}}}\\ V_{\text{TV}}=\frac{\left(120\text{ V}\right)\times \left(52\Omega \right)}{\left(2.5\Omega \right)+\left(52\Omega \right)}\\ V_{\text{TV}}=115\text{ V}\end{array}$

Conclusion:

Thus, the voltage drop across the television is $115\text{ V}$ .

(c)

To determine

The equivalent resistance of the two appliances.

Answer to Problem 67A

The equivalent resistance of the two appliances is $9.8\Omega$ .

Explanation of Solution

Given:

The power of the television is $P=275\text{ W}$ .

The voltage is $V=120\text{ V}$ .

The resistance of the hair dryer is $R_{\text{dryer}}=12\text{Ω}$ .

Calculation:

Refer part (a).

The resistance of the television is $R_{\text{TV}}=52\Omega$ .

The equivalent resistance is,

$\begin{array}{l}\frac{1}{R}=\frac{1}{R_{\text{TV}}}+\frac{1}{R_{\text{dryer}}}\\ R=\frac{1}{\left(\frac{1}{R_{\text{TV}}}+\frac{1}{R_{\text{dryer}}}\right)}\\ R=\frac{1}{\left(\frac{1}{52\Omega }+\frac{1}{12\Omega }\right)}\\ R=9.8\Omega \end{array}$

Conclusion:

Thus, the equivalent resistance of the two appliances is $9.8\Omega$ .

(d)

To determine

The potential difference across the television and the hair dryer.

Answer to Problem 67A

The potential difference across the television and the hair dryer is $96\text{ V}$ .

Explanation of Solution

Given:

The power of the television is $P=275\text{ W}$ .

The voltage is $V=120\text{ V}$ .

The resistance of the wire is $R_{\text{wire}}=2.5\text{Ω}$ .

Formula used:

The expression for the voltage drop is,

$V_{\text{B}}=\frac{V{R}_{\text{B}}}{R_{\text{A}}+R_{\text{B}}}$

Here, $V$ is the voltage, $R_{\text{A}}$ is the resistance and $R_B$ is the resistance.

Calculation:

Refer part (c).

The equivalent resistance of the two appliances is $R=9.8\Omega$ .

The potential difference across the television and the hair dryer is,

  $\begin{array}{l}{V}_{\text{1}}=\frac{VR}{R_{\text{wire}}+R}\\ V_{\text{1}}=\frac{\left(120\text{ V}\right)\times \left(\text{9}\text{.8 }\Omega \right)}{\left(2.5\Omega \right)+\left(\text{9}\text{.8 }\Omega \right)}\\ V_1=96\text{ V}\end{array}$

Conclusion:

Thus, the potential difference across the television and the hair dryer is $96\text{ V}$ .