If a light bulb is wired in series with a 133-ohm resistor and they are connected across a 131-volt source and the power delivered to the bulb is 22.9 watts then what are the two possible resistances?

Answer 1

First, we assume you are asking for the resistance of the bulb. Second, there can be only one possible resistance. The bulb has one and only one resistance. The equation for solving this problem may be a quadratic or even a cubic equation, which may have multiple roots, but only ONE can be a real-world answer. Third, let's list everything we know about the circuit. Let's label the power source Vcc. We'll label the known resistor R, and the resistance of the bulb Rb. We'll label the power dissipated by the bulb as simply P. We'll also label the current flowing through the bulb as I. What do we know about power? The power dissipated by a resistor is given by three well-known formulas: P = IV, P = I2R, or P = V2/R. (The second and third are derived from the first.) The second one will come in handy for us right now. We can write: [Equ. 1] P = I2Rb What else do we know? Since it's a series circuit, the current flowing through the bulb, I, is the same as the current flowing through the resistor. And we know from Ohm's Law that I = V/RT, where RT is the sum of the two resistances in the circuit: R + Rb. So, we can write: [Equ. 2] I = Vcc/(R + Rb) So, substituting for I in Equ. 1 with the I we found in Equ. 2 we have: P = [Vcc/(R + Rb)]2Rb = [Vcc2/(R + Rb)2]Rb = Vcc2Rb/(R2 + 2RRb + R2b) Phew! The algebra's getting a bit hairy. We need to do a few more manipulations. We can swap the P on the left side of the equal sign with the denominator of the fraction on the right side of the equal sign, so we write: R2 + 2RRb + R2b = Vcc2Rb/P We now multiply both sides of the equation by P and we write: PR2 + 2PRRb + PR2b = Vcc2Rb Stay with me now; we're almost there. If we subtract Vcc2Rb from both sides and gather similar terms, we can write: PR2b+ (2PR - Vcc2)Rb + PR2 = 0 We can now substitute all the known values into the equation above. We know Vcc = 131, R = 133, and P = 22.9. So, we can write: 22.9R2b - 11070Rb + 405078 = 0 So, we have quadratic equation, which will have two roots for Rb. Using the quadratic formula, we determine that there are two real roots: Rb = 443.5 and 39.9. The second result is thrown out since it is not consistent with Equ. 1 and Equ. 2.