A light bulb is wired in series with a 142 resistor, and they are connected across a 120 V source. The power delivered to the light bulb is 22.3 W. What are the two possible resistances of the light bulb?
P = 142I^2 = Power to the resistor.
P = 22.3W. = Power to bulb.
P = 120I = Power from source.
The total power delivered is = to the
power from the source:
142I^2 + 22.3 = 120I,
142I^2 - 120I + 22.3 = 0,
Use Quadratic Formula and get:
I = 0.56914A; I = 0.27593A.
P = (0.56914)^2 * R = 22.3,
R = 22.3 / (0.56914)^2 = 68.8 Ohms.
R = 22.3 / (0.27593)^2 = 292.9 Ohms.
To find the resistances of the light bulb, we can use the power formula:
Power (P) = (Voltage)^2 / Resistance
Given that the power (P) is 22.3 W and the voltage (V) is 120 V, we can rearrange the formula to solve for resistance (R):
Resistance (R) = (Voltage)^2 / Power
Substituting the given values:
Resistance (R) = (120)^2 / 22.3
Calculating this:
Resistance (R) ≈ 646.25 Ω
Since the light bulb is wired in series with a 142 Ω resistor, the total resistance of the circuit is the sum of the resistance of the light bulb and the 142 Ω resistor.
Total Resistance (R_total) = Resistance of Light Bulb + Resistance of 142 Ω Resistor
From this, we can find the two possible resistances of the light bulb:
1. R_total = R_lightbulb + 142
646.25 = R_lightbulb + 142
R_lightbulb ≈ 504.25 Ω
2. R_total = 142 + R_lightbulb
646.25 = 142 + R_lightbulb
R_lightbulb ≈ 504.25 Ω
Therefore, the two possible resistances of the light bulb are approximately 504.25 Ω.