An electric current of 50 +/- 1 mA flows through an electric bulb filament that has a resistance of 10 +/- 2 ohm. Estimate the power dissipated by the filament and the associated error in the power.
The power dissipated is
I^2*R = (.05^2)(10) = 0.025 W
That is pretty feeble for a light bulb.
The relative current error is negligible compared to the relative resistance error of 20%.
The power measurement uncertainty is therefore 20% or +/- 0.005 W
To estimate the power dissipated by the filament, we can use the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance.
First, let's calculate the power using the nominal values of the current and resistance:
P = (50 mA)^2 * 10 ohm = 2500 * 10^-6 A^2 * 10 ohm = 2500 * 10^-5 A^2 * ohm = 2500 * 10^-5 W ≈ 0.025 W
Now, let's calculate the error in power. To do this, we can use the formula for propagating the error in a product of variables. In our case, the error in power ΔP can be calculated as follows:
ΔP = sqrt((dP/dI * ΔI)^2 + (dP/dR * ΔR)^2)
where ΔI is the error in current, ΔR is the error in resistance, and dP/dI and dP/dR are the partial derivatives of power with respect to current and resistance, respectively.
Given that ΔI = 1 mA and ΔR = 2 ohm, we need to calculate dP/dI and dP/dR.
Taking the partial derivative of power with respect to current, we have:
dP/dI = 2 * I * R
Taking the partial derivative of power with respect to resistance, we have:
dP/dR = I^2
Now, let's substitute the values into the error formula:
ΔP = sqrt((2 * I * R * ΔI)^2 + (I^2 * ΔR)^2)
= sqrt((2 * 50 * 10^-3 A * 10 ohm * 1 mA)^2 + (50 mA)^2 * (2 ohm)^2)
= sqrt((0.1 A * 10 ohm)^2 + (50 * 10^-3 A)^2 * (2 ohm)^2)
= sqrt((1 ohm)^2 + (0.05 ohm)^2)
≈ sqrt(1.01 ohm^2)
≈ 1 ohm
Thus, the associated error in the power is approximately 1 watt.
Therefore, the power dissipated by the filament is estimated to be 0.025 W with an associated error of approximately 1 W.