If a 12V product is dissipating 50W, what is the load resistance it represents?

If a 12V product is dissipating 50W, what is the load resistance it represents?

• A. 3.6 ohms
• B. 4.8 ohms
• C. 2.9 ohms
• D. 6.0 ohms

Answer: (C)

To determine the load resistance of a product dissipating 50W at a voltage of 12V, we can use Ohm’s law and the power formula. The power (P) dissipated in a resistor can be expressed in terms of voltage (V) and resistance (R) using the formula: \[ P = \frac{V^2}{R} \] Rearranging this formula to solve for resistance gives us: \[ R = \frac{V^2}{P} \] Substituting the given values into the formula: - V = 12V - P = 50W Calculating the resistance: \[ R = \frac{(12V)^2}{50W} \] \[ R = \frac{144V^2}{50W} \] \[ R = 2.88 \, \text{ohms} \] When rounded, this value is approximately 2.9 ohms, making it the correct representation of the load resistance that the 12V product exhibits when dissipating 50W. Understanding this calculation is essential for electronic technicians, as it facilitates the analysis of power and resistance in electrical circuits.

To determine the load resistance of a product dissipating 50W at a voltage of 12V, we can use Ohm’s law and the power formula. The power (P) dissipated in a resistor can be expressed in terms of voltage (V) and resistance (R) using the formula:

[ P = \frac{V^2}{R} ]

Rearranging this formula to solve for resistance gives us:

[ R = \frac{V^2}{P} ]

Substituting the given values into the formula:

• V = 12V

• P = 50W

Calculating the resistance:

[ R = \frac{(12V)^2}{50W} ]

[ R = \frac{144V^2}{50W} ]

[ R = 2.88 , \text{ohms} ]

When rounded, this value is approximately 2.9 ohms, making it the correct representation of the load resistance that the 12V product exhibits when dissipating 50W. Understanding this calculation is essential for electronic technicians, as it facilitates the analysis of power and resistance in electrical circuits.