Chapter 23, Problem 81A

(a)

To determine

To compute: the equivalent resistance of the circuit consisting of just the light and the lead lines to and from the light.

Answer to Problem 81A

The equivalent resistance of the circuit containing the lamp and the load lines is $\boxed{R_{\text{equi}}=240.5\Omega }$ .

Explanation of Solution

Given:

Resistance of the wires of the kitchen light is $25\Omega$ .

The resistance of the light is 0.24 $\text{kΩ}$ .

Potential difference of switch box is 120 V.

Formula used:

The given diagram is shown below:

  

The lamp and the two load lines are connected in series. The equivalent resistance of the circuit having the lamp and two load lines is equal to the sum of the resistance of the lamp and two load lines that is,

  $R_{\text{equi}}=R_{\text{lamp}}+R_{\text{load}\text{line}}+R_{\text{load}\text{line}}$

  $R_{\text{equi}}=R_{\text{lamp}}+2R_{\text{load}\text{line}}$  ......(1)

Calculation:

Substitute the values of $R_{lamp}$ and $R_{load}$ in equation (1)

  $\begin{array}{l}{R}_{\text{equi}}=240+2\left(0.25\right)\\ R_{\text{equi}}=240+0.50\\ \boxed{R_{\text{equi}}=240.5\Omega }\end{array}$

Conclusion:

Hence, the equivalent resistance of the circuit containing the lamp and the load lines is $\boxed{R_{\text{equi}}=240.5\Omega }$ .

(b)

To determine

the current through the light.

Answer to Problem 81A

The current in the lamp is $i=0.5\text{A}$ .

Explanation of Solution

Given:

Resistance of the wires of the kitchen light is $25\Omega$ .

The resistance of the light is 0.24 $\text{kΩ}$ .

Potential difference of switch box is 120 V.

The equivalent resistance of the circuit containing the lamp and the load lines is $R_{\text{equi}}=240.5\Omega$ .

Formula used:

The given diagram is shown below:

  

The current in the lamp is equal to the voltage across it divided by the equivalent resistance of the circuit containing the lamp and the two load lines is

  $i=\frac{V}{R_{\text{equi}}}$  ......(1)

Here, V is the voltage across the lamp.

Calculation:

As the lamp is connected in parallel to the switch box, the voltage across the lamp is equal to the voltage to the switch box. Using this, the above equation becomes,

  $i=\frac{V_{\text{switch}\text{box}}}{R_{\text{equi}}}$  ......(2)

Substitute the values of $V_{\text{switch}\text{box}}$ and $R_{\text{equi}}$ in equation (2)

  $\begin{array}{l}i=\frac{120\text{V}}{240.5\Omega }\\ \boxed{i=0.5\text{A}}\end{array}$

Conclusion:

Hence, the current in the lamp is $i=0.5\text{A}$ .

(c)

To determine

the power used in the light.

Answer to Problem 81A

The power dissipated in the lamp is $\boxed{P=60\text{W}}$ .

Explanation of Solution

Given:

Resistance of the wires of the kitchen light is $25\Omega$ .

The resistance of the light is $0.24\text{kΩ}$ .

Potential difference of switch box is 120 V.

The equivalent resistance of the circuit containing the lamp and the load lines is $R_{\text{equi}}=240.5\Omega$ .

Formula used:

The given diagram is shown below:

  

The power dissipated in the lamp is given by the formula,

  $P=Vi$  ......(1)

Calculation:

Substitute the values of $V$ and $i$ in the equation (1)

  $\begin{array}{c}P=120\times 0.5P\\ \boxed{P=60\text{W}}\end{array}$

Conclusion:

Hence, the power dissipated in the lamp is $\boxed{P=60\text{W}}$ .