You have a 10 kΩ resistor that is rated to withstand a maximum power of 4 W. If you connect this resistor with

Question

If you have a 10 kΩ resistor that is rated to withstand a maximum power of 4 W. If you connect this resistor within a DC circuit, what is the highest voltage that can be safely applied across this resistor, and what is the corresponding maximum current that can flow through the resistor?

Answer 1

P=V^2/R
sqrt (4*10K) = V

I = V/R

Answer 2

According to ohms law, Voltage (V) = Current (I) * Resistance (R) and Power (P) = Voltage (V) * Current (I). Combining these two formulas, you get P = V^2 / R (the square of the voltage over the resistance). Or, V = sqrt(P*R).

So, your answer is this:

Highest Voltage = sqrt(4*10k) = 200 volts
Maximum Current = 200/10k = 20 mA

If you apply more voltage than this, but the resistor will heat up and the resistance will begin to change. Also note that the resistor has a tolerance that should be taken into consideration. And resistors actually change value depending on the outside temperature too.

Imagine what would happen if your resistor had 5% less resistance, was in a non ventilated box, and started to decrease its resistance as it got hotter. As a rule of thumb, it is best to get twice the wattage rating as you need.

Answer 3

Power=I^2R=4kw=4000w
I^2=4000/10=400
I=20amps
Rated current is 20 amps
Safe value of DC voltage
=10 Ohms*20amps=200 volts
Current flowing=20 amps

Answer 4

i will attempt to respond to this one devoid of any formulation. a watt volume is computed by skill of multiplying voltage cases amps. you get the amp volume by skill of taking voltage divided by skill of ohms. So lets round you resistor volume to 100 only to make the reason a lot less complicated: with 10 volts (also rounded #), divided by skill of 100 resistance, you recognize that the amps of this circuit is .1A So now together with your voltage and amperage numbers figgured out, you could now decide the wattage: 10 volts cases .a million amps = a million watt. it truly is how "ohms regulation" works and converts all electric powered measurements...they are interchangeable, and linked mathematically. With this hassle-free understanding now conventional, you ought to be able to decide your question on your own distinctly honestly. I recommend doing the maths your self, in spite of the indisputable fact that the answer is 12 volts pushing by that resistor max earlier it burns out. :)

Answer 5

P= V^2/R
V^2 = PR
V = sqrt PR = sqrt 10^3 X 4=10sqrt40 = 63.2 Volts
P = I^2R
I^2 = P/R
I = sqrt P/R = sqrt 4/10^3 = .0632 A or 63.2 mA