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 15 kΩ resistor is 5.0 W, what is the maximum allowable potential difference across the terminals of the resistor?

(b) A 9.0 kΩ resistor is to be connected across a 120-V potential difference. What power rating is required?

(c) A 100.0 Ω and a 150.0 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 resistor under these conditions?

Short answer

(a)The maximum allowable potential difference across the terminals of the resistor is 274 V.

(b)When a 9.0 kΩ resistor is to be connected across a 120-V potential difference, the power rating that is required is 1.60 W

(c)The greatest this potential difference can be without overheating either resistor is 28.8 V the rate of heat generated in each resistor R₁₅₀ is 1.32 W and R₁₅₀ is 2.0 W.

Step-by-step solution

Given data

• R = 15 kΩ
• P = 5.0 W

The maximum allowable potential difference across the terminals of the resistor

a)

The dissipated power in the resistor is related to the voltage between the terminals of the resistor as given:

$$\begin{gathered} \mathit{P}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}} \\ \mathit{V}\mathbf{=}\sqrt{\mathbf{P}\mathbf{R}} \end{gathered}$$

Here V is the voltage drop, R is the resistance of the resistor, and P is the power dissipation.

Substitute the values to get V as:

$$\begin{gathered} V=\sqrt{PR} \\ =\sqrt{\left(5W\right)/\left(15000\Omega \right)} \\ =274.0V \end{gathered}$$

The maximum allowable potential difference across the terminals of the resistor is 274 V.

Power rating

b)

Given data

• R=9 kΩ
• V=5.0 V

The dissipated power in the resistor is related to the voltage between the terminals of the resistor as given :

$$\begin{gathered} P=\frac{V^2}{R} \\ =\frac{{\left(120V\right)}^2}{9000\Omega } \\ =1.60W \end{gathered}$$

When a 9.0 kΩ resistor is to be connected across a 120-V potential difference, the power rating that is required is 1.60 W.

The greatest this potential difference can be without overheating either resistor, and the rate of heat generated is in each resistor under these conditions

c)

• Two resistors R₁ = 100.0 Ω and R₂ = 150.0 Ω are connected in series.

The potential V across the two resistors and the rated heat P for each one. The given power P = 2 W belongs to the greater resistor, where the power here is the maximum loss, and the maximum loss will belong to the greatest resistance so the rated heat due to R₂ is

$P_2=2W$

Both resistors are in series, this means the current is the same for both resistors and the total voltage is the sum of the individual resistors. Now, find the current I then calculate the voltage by using Ohm's law.

Use the value of P, to find the current I, where the dissipated power is related to the current by the equation as:

role="math" localid="1664191637037" $$\begin{gathered} P=I^{2}R \\ I=\sqrt{\frac{P}{R}} \\ l=\sqrt{P_2/R_2} \\ =\sqrt{\left(2W\right)/\left(150.0\Omega \right)} \\ =0.115A \end{gathered}$$

This current is the current through the circuit and the voltage across both resistors could be calculated by Ohm's law as:

$$\begin{gathered} V=l\left(R_1+R_2\right) \\ =0.115A\left(100.0\Omega +150.0\Omega \right) \\ =28.8V \end{gathered}$$

Now get the rated heat of R=

$$\begin{gathered} P_1=l^2{R}_2 \\ ={\left(0.115A\right)}^2\left(100.0\Omega \right) \\ =1.32W \end{gathered}$$

The greatest this potential difference can be without overheating either resistor is 28.8 V the rate of heat generated in each resistor R₁₀₀ is 1.32 W and R₁₅₀ is 2.0 W.