Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. (a) If the power rating of a [email protected] Ω resistor is 5.0 W, what is the maximum allowable potential difference across the termi- nals of the resistor? (b) A [email protected]Ω resistor is to be connected across a 120-V potential difference. What power rating is required? (c) A [email protected]Ω and a [email protected]Ω resistor, both rated at 2.00 W, are connected in series across a variable potential difference. What is the greatest this potential difference can be without overheating either resistor, and what is the rate of heat generated in each resis- tor under these conditions?

Answer 1

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$, and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$, then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

$V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V$

(b) 1.6 W

In this case, we have:

$R=9.0 k\Omega = 9000 \Omega$ is the resistance of the resistor

$V=120 V$ is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

$P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W$

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

$R_1 = 100 \Omega\\R_2 = 150 \Omega$

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

$V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V$

While the greatest potential difference allowed in the second resistor is

$V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V$

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

$P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W$

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).