A light bulb is wired in series with a 133-Ω resistor, and they are connected across a 120-V source. The power delivered to the light bulb is 22.5 W. What are the two possible resistances of the light bulb?
Power to bulb= I^2 R
but I= 120/(R+133)
22.5= 120^2/(R+133)^2 * R
solve for R. Notice it is a quadratic.
Joe, check your 1-24-11, 5:19pm Post.
((R1)(V^2))/((R1+R2)^2)=Power (watts)
in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)
To find the two possible resistances of the light bulb, we need to use the formulas for power and resistance in a series circuit.
The formula for power in a circuit is:
P = V^2 / R
where P is power in watts, V is voltage in volts, and R is resistance in ohms.
We are given that the power delivered to the light bulb is 22.5 W, the voltage across the circuit is 120 V, and the resistor connected in series with the light bulb has a resistance of 133 Ω.
We can rearrange the formula for power to solve for resistance:
R = V^2 / P
Let's substitute the given values:
R = (120 V)^2 / 22.5 W
R = 14400 V^2 / 22.5 W
R ≈ 640 Ω
So one possible resistance of the light bulb is approximately 640 Ω.
However, since we are asked for two possible resistances, we need to consider the two different scenarios when calculating the resistance. In a series circuit, the total resistance is equal to the sum of the individual resistances. Therefore, we can calculate the second possible resistance by subtracting the known resistor's resistance from the total resistance of the circuit.
Total resistance = known resistor's resistance + light bulb's resistance
Given:
Total resistance = 133 Ω (known resistor's resistance)
One possible resistance = 640 Ω
Second possible resistance = Total resistance - known resistor's resistance
Second possible resistance = 640 Ω - 133 Ω
Second possible resistance ≈ 507 Ω
Therefore, the two possible resistances of the light bulb are approximately 640 Ω and 507 Ω.