Question

Q5 A 0.27 k resistor is rated 2W. Compute the maximum voltage that can be applied and the maximum current that it can carry without exceeding its rating, (15)

Answer 1

The maximum voltage that can be applied to a 0.27 kΩ resistor is approximately 2.62 V, and the maximum current it can carry without exceeding its rating is approximately 9.7 mA.

Step-by-step explanation:

To calculate the maximum voltage that can be applied to a 0.27 kΩ resistor rated 2W, we can use the formula P = V²/R, where P represents the power, V is the voltage, and R is the resistance.

Substituting the values into the formula, we get 2 = V²/0.27, which can be rearranged to V² = 2 * 0.27. Solving this equation, we find V ≈ 2.62 V. Therefore, the maximum voltage that can be applied to the resistor without exceeding its rating is approximately 2.62 V.

To calculate the maximum current that the resistor can carry, we can use Ohm's Law, which states that I = V/R, where I represents the current, V is the voltage, and R is the resistance. Substituting the values, we get I = 2.62/0.27. Solving this equation, we find I ≈ 9.7 mA. Therefore, the maximum current that the resistor can carry without exceeding its rating is approximately 9.7 mA.

Answer 2

To find the maximum voltage and current for a 0.27 k resistor with a 2W rating, use the formulas P = V^2 / R and P = I^2 * R respectively. The maximum voltage is √(2W * 0.27 k) volts and the maximum current is √(2W / 0.27 k) amperes.

Step-by-step explanation:

To determine the maximum voltage that can be applied to the 0.27 k resistor without exceeding its rating, we need to use the formula P = V^2 / R, where P is the power rating of the resistor, V is the voltage, and R is the resistance.

Substituting the given values, we have 2W = V^2 / 0.27 k.

Solving for V, we get V = √(2W * 0.27 k).

Similarly, to compute the maximum current that the resistor can carry without exceeding its rating, we use the formula P = I^2 * R, where P is the power rating of the resistor, I is the current, and R is the resistance.

Substituting the given values, we have 2W = I^2 * 0.27 k.

Solving for I, we get I = √(2W / 0.27 k).

Therefore, the maximum voltage that can be applied to the resistor without exceeding its rating is approximately √(2W * 0.27 k) volts, and the maximum current it can carry is approximately √(2W / 0.27 k) amperes.