Chapter 23, Problem 63A

(a)

To determine

the resistance of the light string.

Answer to Problem 63A

The total resistance of the light string is 225 $\Omega$ .

Explanation of Solution

Given:

Total power dissipation is $P=64\text{W}$ .

  $V_{\text{source}}=120\text{V}$ .

Formula used:

Total power dissipation can be found by using formula

  $P=\frac{V^2}{R}$  ......(1)

Here, $V$ and $R$ are potential difference of the source and equivalent resistance of the circuit.

Calculation:

Rearrange the equation (1) for R,

  $R=\frac{V^2}{P}$  ......(2)

Substitute the values of $V$ and $P$ in equation (2)

  $\begin{array}{l}R=\frac{120\text{V}}{64\text{W}}\\ \boxed{R=225\Omega }\end{array}$

Conclusion:

Hence, the total resistance of the light string is 225 $\Omega$ .

(b)

To determine

the power used by string.

Answer to Problem 63A

The resistance of a single light will be $12.5\Omega$ .

Explanation of Solution

Given:

Total power dissipation is $P=64\text{W}$ .

  $V_{\text{source}}=120\text{V}$ .

Formula used:

Total power dissipation can be found by using formula

  $P=\frac{V^2}{R}$  ......(1)

Here, $V$ and $R$ are potential difference of the source and equivalent resistance of the circuit.

Calculation:

It is given that, the light string is made of 18 identical lights connected in series.

In series circuit, the equivalent resistance is the sum of the resistance of the entire resistor connected in it. Since, the light are identical the resistance of the entire 18 resistor will be same.

Therefore, the resistance of a single light will be $\frac{225\Omega }{18}=12.5\Omega$ .

Conclusion:

Hence, the resistance of a single light will be $12.5\Omega$ .

(c)

To determine

whether power decreases or increases when the bulb burned out.

Answer to Problem 63A

The power will decrease.

Explanation of Solution

Given:

Total power dissipation is $P=64\text{W}$ .

  $V_{\text{source}}=120\text{V}$ .

Resistance of a single light is $R=12.5\Omega$ .

Formula used:

Power can be expressed as

  $P=I^{2}R$  ......(1)

Current can be expressed as

  $I=\frac{V}{R}$  ......(2)

Calculation:

Substitute the values,

  $\begin{array}{l}I=\frac{120\text{V}}{225\Omega }\\ I=0.533\text{A}\end{array}$

The current in the circuit is 0.533 A.

Now, substitute the values of $I$ and R in equation (1)

  $\begin{array}{l}P={\left(0.533\text{A}\right)}^2\times \left(12.5\Omega \right)\\ \boxed{P=3.56\text{W}}\end{array}$

Since this power is less than the total power, so power will decrease.

Conclusion:

Hence, the power will decrease.