Question

1. the maximum voltage is 1 v and the maximum current is 1 a in a given resistor. should a 1 w or a 2 w resistor be used? why?
2. a 6.8 kω resistor burned out in a circuit. you must replace it with another resistor with the same resistance value. if the resistor carries 10 ma, what should its power - rating be? assume that you have available resistors in all the standard power ratings.
3. a certain type of power resistor comes in the following ratings: 3 w, 5 w, 8 w, 12 w, 20 w. your particular application requires a resistor that can handle approximately 8 w. which rating would you use for a minimum safety margin of 20% above the rated value? why?

Explanation:

Step1: Recall power - resistance - current - voltage relationship

The power formula is $P = VI=I^{2}R=\frac{V^{2}}{R}$.

Step2: Calculate power for the first resistor

Given $V = 1V$ and $I = 1A$, using $P=VI$, we have $P=(1V)\times(1A) = 1W$. So a 1 - W resistor should be used as the power dissipated in the resistor is 1W.

Step3: Calculate power for the second resistor

The resistance $R = 6.8k\Omega=6800\Omega$ and current $I = 10mA = 0.01A$. Using $P = I^{2}R$, we get $P=(0.01A)^{2}\times6800\Omega=0.68W$. Looking at standard power ratings, the next - higher standard power rating should be chosen.

Step4: Calculate power for the third resistor

If a minimum safety margin of 20% is required for a resistor that can handle approximately 8W, let the required power rating be $P_{r}$. Then $P_{r}=(1 + 0.2)\times8W=9.6W$. So the 12 - W rating should be chosen.

Answer:

1. A 1 - W resistor should be used since $P = VI=(1V)\times(1A)=1W$.
2. First, calculate $P = I^{2}R=(0.01A)^{2}\times6800\Omega = 0.68W$. The next - higher standard power rating should be chosen.
3. The required power rating is $(1 + 0.2)\times8W=9.6W$, so the 12 - W rating should be chosen.