Problem 20 - Basic Engineering Mathematics

Question

The power \(P\) watts dissipated in an electrical circuit may be expressed by the formula \(P=\frac{V^{2}}{R}\). Evaluate the power, correct to 3 significant figures, given that \(V=17.48 \mathrm{~V}\) and \(R=36.12 \Omega\).

Step-by-Step Solution

Answer

The power is approximately 8.46 watts.

Step 1: Understand the Formula

The formula used to calculate power in terms of voltage and resistance is \( P = \frac{V^2}{R} \). Here, \( P \) is the power in watts, \( V \) is the voltage in volts, and \( R \) is the resistance in ohms.

Step 2: Substitute Known Values

Insert the given values into the formula: \( V = 17.48 \mathrm{~V} \) and \( R = 36.12 \Omega \). Hence, the equation becomes \( P = \frac{(17.48)^2}{36.12} \).

Step 3: Calculate Voltage Squared

Calculate \( V^2 \). With \( V = 17.48 \), we have \( 17.48^2 = 305.7104 \).

Step 4: Divide by the Resistance

Now, divide \( 305.7104 \) by \( 36.12 \) to find \( P \). Thus, \( P = \frac{305.7104}{36.12} \approx 8.463 \).

Step 5: Round to Three Significant Figures

Round \( 8.463 \) to three significant figures. The result is \( 8.46 \).

Key Concepts

Power Formula

The Power Formula is a fundamental concept in electrical circuits. It helps us figure out how much electrical power is being used in a circuit. Power, denoted as \( P \), is expressed in watts (W), serving as a measure of the rate at which energy is used or converted.

The formula used for power in relation to voltage and resistance is:

• \( P = \frac{V^2}{R} \)

Here:

• \( V \) stands for voltage in volts (V)
• \( R \) represents resistance in ohms (Ω)

This formula essentially tells us that power is proportional to the square of the voltage. Thus, a larger voltage will lead to considerably more power, provided the resistance stays constant.

If you know the voltage across a resistor and the resistance, you can quickly calculate the power by plugging these values into the equation. Understanding this relationship is essential in designing circuits to ensure they transfer or consume energy efficiently without overheating or underperforming.

Voltage Calculation

Calculating voltage across different components in a circuit is a crucial skill in understanding how electrical systems function. In our given formula for power, voltage is squared, highlighting how influential it is in determining power.

When dealing with circuits, voltage can be thought of as the "push" that helps drive the electrical current through a circuit.

The formula given, \( V = \sqrt{P \times R} \), arises from rearranging our original power equation, where if you know the values of power and resistance, you can deduce the voltage. However, in scenarios like the example, voltage is already known:

• The original value provided was \( V = 17.48 \text{ V} \).

Calculating \( V^2 \) from this value involves simple arithmetic, \( 17.48^2 \), which yields \( 305.7104 \text{ V}^2 \). This squared value plays directly into power calculation, as it shows how increases in voltage can dramatically escalate the resulting power.

Always remember; small differences in voltage can lead to significant changes in the power output.

Resistance in Ohms

Resistance is essentially what opposes the flow of electrical current in a circuit. It is measured in ohms (Ω), and the formula for power directly incorporates resistance to show how it influences power consumption.

Resistance can be visualized as a kind of friction for electrons, slowing down the current. The power formula, \( P = \frac{V^2}{R} \), indicates that:

• As resistance increases, power decreases, provided voltage remains constant.
• Conversely, lower resistance at the same voltage level means higher power.

In our earlier example, the resistance value of \( R = 36.12 \Omega \) plays a crucial role in determining power. Using this resistance, the squared voltage, and the provided formula, we can calculate the exact power in watts.

• This calculation results in \( P = \frac{305.7104}{36.12} \approx 8.463 \text{ W} \).

Here we see that even moderate changes in resistance can have a substantial impact on the circuit's power. Accurate resistance measurements are vital for ensuring a circuit's efficiency and safety.