Question

Heat produced in a current carrying conducting wire is \(\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}\) it percentage error in \(\mathrm{I}, \mathrm{R}\) and \(\mathrm{t}\) is \(2 \%, 4 \%\) and \(2 \%\) respectively then total percentage error in measurement of heat energy \(\ldots \ldots \ldots\) (a) \(8 \%\) (b) \(15 \%\) (c) \(5 \%\) (d) \(10 \%\)

Short answer

The total percentage error in the measurement of heat energy is approximately 6%. The closest answer among the provided options is (c) 5%.

Step-by-step solution

Identify the given variables and their percentage errors

The given variables and their percentage errors are as follows: - I (current) has a 2% error - R (resistance) has a 4% error - t (time) has a 2% error

Determine the error formula for heat energy

To find the total percentage error in the heat energy produced, we need to use the formula for calculating the percentage error in a derived quantity, based on the individual percentage errors of the variables involved. The formula for percentage error in a derived quantity is given by: \[ \% \text{error} = \sqrt{(\frac{\partial (Derived \, Quantity)}{\partial(Variable)})^2 \times (\% \, error\, in \, variable)^2} \] In our case, the derived quantity is H (heat energy), and the variables are I, R, and t.

Apply the error formula to the given equation

Now let's apply the formula for percentage error to the given equation for heat energy produced, H = I^2Rt. \[ \% \text{error in H} = \sqrt{(\frac{\partial H}{\partial I})^2 \times (\% \text{error in I})^2 + (\frac{\partial H}{\partial R})^2 \times (\% \text{error in R})^2 + (\frac{\partial H}{\partial t})^2 \times (\% \text{error in t})^2} \] Next, we calculate the partial derivatives of H with respect to I, R, and t: 1. \(\frac{\partial H}{\partial I} = 2I \times Rt\) 2. \(\frac{\partial H}{\partial R} = I^2 \times t\) 3. \(\frac{\partial H}{\partial t} = I^2 \times R\)

Substitute the given errors and partial derivatives into the error formula

Now, substitute the partial derivatives and percentage errors into the formula: \[ \% \text{error in H} = \sqrt{(2I \times Rt)^2 \times (0.02)^2 + (I^2 \times t)^2 \times (0.04)^2 + (I^2 \times R)^2 \times (0.02)^2 } \]

Simplify the expression

We can simplify the expression further by noticing that \(I^2 \times R \times t\) is equal to H; therefore, \[ \% \text{error in H} = \sqrt{(2H)^2 \times (0.02)^2 + (H)^2 \times (0.04)^2 + (H)^2 \times (0.02)^2} \] Divide all the terms by H^2 \[ \% \text{error in H} = \frac{1}{H}\sqrt{4H^2 \times (0.02)^2 + H^2 \times (0.04)^2 + H^2 \times (0.02)^2} \]

Calculate the percentage error

Cancel out H in numerator and denominator \[\% \text{error in H} = \sqrt{4\times(0.02)^2 + (0.04)^2 + (0.02)^2} \] Calculating the values, \[\% \text{error in H} = \sqrt{0.0016 + 0.0016 + 0.0004} = \sqrt{0.0036} \] The final percentage error in H is, \[\% \text{error in H} = 0.06 \] Converting it back to percentage, we get \(0.06 \times 100 = 6 \% \) error in H The closest answer is: (c) 5%

Key concepts

Error Propagation

In the context of scientific measurements, error propagation is a vital concept that explains how uncertainties in measurements influence the uncertainty of a derived quantity. In our example, we are examining how errors in individual measurements of current, resistance, and time propagate when calculating thermal energy in a current-carrying wire using the formula \[ H = I^2Rt \].When considering percentage errors, error propagation helps us determine the overall uncertainty in a derived quantity by combining the percentage uncertainties of the individual measured quantities. This is achieved by taking derivatives of the equation with respect to each variable. By calculating each variable's contribution to the total error separately and then combining these using the Pythagorean theorem, we arrive at the total percentage error. This process allows us to understand how sensitive the outcome is to errors in each of the variables, guiding us in focusing efforts to reduce uncertainty in our measurements where it matters most.

Thermal Energy

Thermal energy refers to the energy that is generated and measured by heat. In the context of electrical circuits, when a current flows through a conductor, it can produce heat, or thermal energy, due to the resistance present in the material. This phenomenon is described by the formula \[ H = I^2Rt \],where \(H\) is the thermal energy, \(I\) is the current, \(R\) is the resistance, and \(t\) is the time for which the current flows.This formula not only helps in calculating the amount of heat generated but is also a key example of how electrical energy is converted into thermal energy. Practical applications can be seen in household appliances like electric heaters, where electricity is converted into heat to warm a space. Understanding this conversion is crucial in fields like electrical engineering and thermal management systems.

Current Electricity

Current electricity is the flow of electric charge through a conductor, usually driven by a potential difference (voltage). In the provided formula \[ H = I^2Rt \],current electricity plays a central role. The current \(I\), as a component of the formula, is squared, highlighting its significant impact on the amount of heat produced by the circuit.Current electricity is fundamental in understanding many aspects of electrical circuits, including power transmission, energy consumption, and heat generation. When calculating the thermal energy produced, the squared relationship in the formula underscores that even small errors in measuring current can have a large effect on the calculated thermal energy.Understanding current flow is essential not only for computing thermal energy but also for designing safe and efficient circuits. Knowing how electricity behaves, interacts with materials, and produces other forms of energy helps in developing advanced technological solutions.