Chapter 19, Problem 14P

To determine

The resistance of each bulb and fraction of total power wasted in the leads.

Answer to Problem 14P

The resistance of each bulb is $445.53\Omega$ and the fraction of power wasted in the leads is $0.088\%$ .

Explanation of Solution

Given info:

Voltage of the source, $V=110\text{V}$

Total resistance of the leads, $R_l=1.6\Omega$

Current trough each bulb, $I_b=240\text{mA}=240\text{mA}\times \frac{1\text{A}}{{10}^3\text{mA}}=0.240\text{A}$

The number of bulbs is, $n=8$ .

The number of leads is, $n_l=2$ .

Formula Used:

The expression to calculate the resistance of each bulb is,

  $R_b=n\left(\frac{V}{n{I}_b}-R_l\right)$

The expression to calculate the fraction of power wasted in the leads is,

  $x=\left(\frac{n_l{R}_l}{n{R}_b}\right)\times 100$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_b=\left(8\right)\left(\frac{110\text{V}}{8\left(0.240\text{A}\right)}-1.6\Omega \right)\\ =445.53\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}x=\left(\frac{2\left(1.6\Omega \right)}{8\left(455.53\Omega \right)}\right)\times 100\\ =0.088\%\end{array}$

Conclusion:

Therefore, the resistance of each bulb is $445.53\Omega$ and the fraction of power wasted in the leads is $0.088\%$ .