Question

A resistor with \(60 \pm 2\, \Omega\) resistance is part of a circuit. The voltage measured across this resistor is \(120 \pm 10\) mV (millivolts). The uncertainty in the estimation of the power dissipated by the resistor is mW (milliwatts). (Round off to one decimal place).

Hint:

Use fractional uncertainty propagation for \(P=V^2/R\): the exponent on each quantity multiplies its fractional error, and independent errors add.

Answer 1

Correct Answer: 0.05

Step 1: Write the power formula in terms of V and R.
Power dissipated by a resistor is \(P = \dfrac{V^2}{R}\).
Here \(V = 120\) mV \(= 0.120\) V with uncertainty \(\delta V = 10\) mV \(= 0.010\) V, and \(R = 60\ \Omega\) with uncertainty \(\delta R = 2\ \Omega\).

Step 2: Find the nominal power.
\[ P = \frac{(0.120)^2}{60} = \frac{0.0144}{60} = 0.00024 \text{ W} = 0.24 \text{ mW} \]

Step 3: Propagate the uncertainty using the power-law rule.
For \(P = V^2 R^{-1}\), taking logarithms gives \(\ln P = 2\ln V - \ln R\).
Differentiating and keeping every error additive (worst-case estimate), the fractional uncertainty in \(P\) is
\[ \frac{\delta P}{P} = 2\frac{\delta V}{V} + \frac{\delta R}{R} \]
The rule says the exponent on a quantity multiplies its fractional uncertainty, and uncertainties from independent quantities add up.

Step 4: Plug in the numbers.
\[ \frac{\delta V}{V} = \frac{0.010}{0.120} = 0.0833, \qquad \frac{\delta R}{R} = \frac{2}{60} = 0.0333 \]
\[ \frac{\delta P}{P} = 2(0.0833) + 0.0333 = 0.1667 + 0.0333 = 0.2 \]

Step 5: Get the absolute uncertainty in power.
\[ \delta P = 0.2 \times 0.24 \text{ mW} = 0.048 \text{ mW} \]
Rounded off, this is about \(0.05\) mW.

Final Answer:
The uncertainty in the power dissipated is about 0.05 mW. \[ \boxed{\delta P \approx 0.05 \text{ mW}} \]