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Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(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).

4 0

3 years ago

find the resistance of a resistor connected to a 3v voltmeter and a 3A ammeter, resistance box along with cells of EMF 3V

yarga [219]

Answer:

0

Explanation:

Emf E= V+ terminal V

3 = 3R + 3

R=0

5 0

3 years ago

You split wood with an ax. How would you change the ax to make splitting the wood easier? You would increase mechanical advantag

alex41 [277]

I think you forgot to give the choices along with the question. I am answering the question based on my knowledge and research. You would increase mechanical advantage by making the blade longer from the cutting edge. I hope that this is the answer that has actually come to your desired help.

6 0

3 years ago

Read 2 more answers

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

kramer

I think the question should be the below:

What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer is the below:

Acceleration .. a = (-) ω² x 
(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) 

x (max) = a(max) /ω² 

x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)

5 0

3 years ago

How is amplitude related to loudness

expeople1 [14]

The loudness of a sound is linked to the size of the vibration which produces it. A big vibration makes a louder sound. Scientists use the word 'amplitude' for the size of waves. For waves on water, it is easy to measure the amplitude.

3 0

3 years ago