Chapter 19, Problem 65GP

(a)

To determine

The method of connection of the filament to obtain mentioned wattage.

Answer to Problem 65GP

The connections of the two filaments should be in parallel connection.

Explanation of Solution

Introduction:

The voltage drop across each device in parallel connections remains same and current trough each device remains same in the series connection of the devices. The defect in one device in the parallel connection does not interrupt the supply for the whole parallel connection but in series connections defect in one device interrupts the supply of the series connection of all devices.

The both filaments should be connected in the parallel connection because the supply to other bulbs do not disturb due to failure of one bulb. In series connection, the failure of one filament disturbs the supply to bulbs because the complete circuit becomes open circuit.

Conclusion:

Thus, the connections of the two filaments should be in parallel connection.

(b)

To determine

The resistance of the each filament.

Answer to Problem 65GP

The resistance of each filament for first bulb is $144\Omega$ , the resistance of each filament for second bulb is $72\Omega$ and the resistance of each filament for third bulb is $48\Omega$

Explanation of Solution

Given info:

The power of first bulb is, $P_1=50\text{W}$ .

The power of second bulb is, $P_2=100\text{W}$ .

The power of third bulb is, $P_3=150\text{W}$ .

The supply voltage for three way bulb is, $V=120\text{V}$ .

The number of filament in bulb is, $n=2$ .

Formula Used:

The expression to calculate the resistance of each filament for first bulb is,

  $R_1=\frac{V^2}{n{P}_1}$

Here,

• n is the number of filament.

The expression to calculate the resistance of each filament for second bulb is,

  $R_2=\frac{V^2}{n{P}_2}$

The expression to calculate the resistance of each filament for third bulb is,

  $R_3=\frac{V^2}{n{P}_3}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_1=\frac{{\left(120\text{V}\right)}^2}{\left(2\right)\left(50\text{W}\right)}\\ =144\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_2=\frac{{\left(120\text{V}\right)}^2}{\left(2\right)\left(100\text{W}\right)}\\ =72\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_3=\frac{{\left(120\text{V}\right)}^2}{\left(2\right)\left(150\text{W}\right)}\\ =48\Omega \end{array}$

Conclusion:

Thus, the resistance of each filament for first bulb is $144\Omega$ , the resistance of each filament for second bulb is $72\Omega$ and the resistance of each filament for third bulb is $48\Omega$ .