Exercise 42 Page 263

We are given an equation for the voltage V required for a circuit. In this equation P represents the power in watts and R represents the resistance in ohms. V=sqrt(PR) There are two circuits: one with a 75-watt light bulb and the other with a 100-watt light bulb. If the resistance of both is 110 ohms, we will find how many more volts are required to light the latter light bulb than the former one. Let's first calculate the number of volts V_1 needed to light the 75-watt light bulb by substituting P_1= 75 and R_1= 110 into the equation.

V_1=sqrt(P_1R_1)

SubstituteII

P_1= 75, R_1= 110

V_1=sqrt(75( 110))

▼

Solve for V_1

Multiply

Multiply

V_1=sqrt(8250)

CalcRoot

Calculate root

V_1=90.829510...

RoundInt

Round to nearest integer

V_1≈ 91

Next, we will find the number of volts V_2 needed to light a 100-watt light bulb. To do so, let's substitute P_2= 100 and R_2= 110 into the equation.

V_2=sqrt(P_2R_2)

SubstituteII

P_2= 100, R_2= 110

V_2=sqrt(100( 110))

▼

Solve for V_2

Multiply

Multiply

V_2=sqrt(11 000)

CalcRoot

Calculate root

V_2=104.880884...

RoundInt

Round to nearest integer

V_2≈ 105

Now that V_1 is about 91 volts and V_2 is about 105 volts, we can approximately calculate the difference between V_2 and V_1. V_2-V_1≈ 105-91 ⇔ V_2-V_1≈ 14 Therefore, about 14 more volts are needed to light a 100-watt light bulb.