Chapter 23, Problem 62A

a.

To determine

Equivalent resistance of light string.

Answer to Problem 62A

Equivalent resistance of the light string $R=225\Omega$ .

Explanation of Solution

Given:

Voltage source, $V=120\text{ V}$

Power used by string, $P=64\text{ W}$

Formula used:

Power absorbed is given by,

  $P=\frac{V^2}{R}$

Where, P is power, v is voltage and R is resistance.

Calculation:

Now, substituting the value of P and V and solve.

  $\begin{array}{l}64=\frac{{120}^2}{R}\\ R=\frac{{120}^2}{64}\\ =225\Omega \end{array}$

Conclusion:

Therefore, equivalent resistance of string light is $225\Omega$ .

b.

To determine

Resistance of single light.

Answer to Problem 62A

Resistance of single light string is, $12.5\Omega$ .

Explanation of Solution

Given:

Voltage source, $V=120\text{ V}$

Power used by string, $P=64\text{ W}$

Formula used:

For series combination equivalent resistance is given by,

  $R=R_1+R_2+R_3+\mathrm{.......}{R}_n$

Where, $R$ is equivalent resistance and $R_1,R_2,R_3\mathrm{.......}{R}_n$ are individual resistances.

Calculation:

From part (a) $R=225\Omega$

Let resistance of each light is $R_1$

Now, substituting the value of R and solve.

  $\begin{array}{c}225=R_1+R_2+R_3+\mathrm{...........}{R}_{18}\\ 225=R_1+R_1+R_1+\mathrm{.............}{R}_1\\ 18R_1=225\\ R_1=\frac{225}{18}\\ =12.5\Omega \end{array}$

Conclusion :

Therefore, resistance of each light string is $12.5\Omega$ .

c.

To determine

Power used by each light.

Answer to Problem 62A

Power of each light is 3.55 W.

Explanation of Solution

Given:

Voltage source, $V=120\text{ V}$

Power used by string, $P=64\text{ W}$

Formula used:

For series combination total power is given by,

  $P=P_1+P_2+P_3+\mathrm{...........}{P}_n$

Where, P is total power and $P_1,P_2,P_3\mathrm{...........}{P}_n$ are individual power of each light.

Calculation:

Let power of each light is $P_1$

Now, substituting the value of P and solve.

  $\begin{array}{c}64=P_1+P_2+P_3+\mathrm{...........}{P}_{18}\\ 64=P_1+P_1+P_1+\mathrm{.............}{P}_1\\ 18P_1=64\\ P_1=\frac{64}{18}\\ =3.55\text{ W}\end{array}$

Conclusion :

Therefore, power used by each light is 3.55 W.