# V = SQRT(P*R) - calculating the resistor voltage rating

#### mishra87

Joined Jan 17, 2016
1,049
Hi All,
I am confused about calculating resistor voltage rating.
part no CRCW0402...C is having 0.063W power and voltage rating is 50V.
https://www.vishay.com/docs/28773/crcwce3.pdf

Let say resistance vale is 100 ohm, what is the maximum voltage this part can handle.

With above formula = V = SQRT(0.063*100) = 2.5V

So I am confused why they have given 50V in their specification.
Lets say i am using 100 in series with some other circuit.

#### dl324

Joined Mar 30, 2015
17,125
So I am confused why they have given 50V in their specification.
That's the maximum working voltage and it's determined by the physical dimensions of the resistor. It has nothing to do with power capability.

This Yaego information for through hole parts shows the relationship for different resistor dimensions:

#### MrChips

Joined Oct 2, 2009
31,079
The power rating and voltage rating are two different things, not related to the P = V*V/R forumula.
The voltage rating is the breakdown voltage.
At 50V, a 100Ω resistor would have to dissipate 25W!

A common thru-hole 1/4W carbon film resistor is rated at 200V or more.

#### mishra87

Joined Jan 17, 2016
1,049
The power rating and voltage rating are two different things, not related to the P = V*V/R forumula.
The voltage rating is the breakdown voltage.
At 50V, a 100Ω resistor would have to dissipate 25W!

A common thru-hole 1/4W carbon film resistor is rated at 200V or more.
100, A 0603 resistor , Power rating = 0.063W, and voltage rating is 50V.
So per you it is going to dissipate 25W.
Still it is not clear to me !
Thanks !

#### Audioguru again

Joined Oct 21, 2019
6,782
Voltage squared/resistance= power (heating).
Then 50V squared / 100 ohms= 25W.
Also, 2.5V squared / 100 ohms= 0.0625W.

#### dl324

Joined Mar 30, 2015
17,125
Still it is not clear to me !
The voltage rating is for that package style and doesn't imply that a 100Ω 0603 resistor can dissipate 25W.

$$\large P=\frac{V^2}{R} -> V=\sqrt{PR} = \sqrt{{63mW}*{100\Omega}}=2.5V$$

The maximum voltage you can have across the resistor is only 2.5V. I hadn't internalized that 0603 resistors were so limiting...

#### Ylli

Joined Nov 13, 2015
1,089
100, A 0603 resistor , Power rating = 0.063W, and voltage rating is 50V.
So per you it is going to dissipate 25W.
Still it is not clear to me !
Thanks !
You are limiting your thinking to steady state conditions. If you put 50 volts across that 0603 resistor, it is going up in smoke. But what if it is the load for a circuit that outputs 50 volts at a 0.1% duty cycle? The resistor will now dissipate only 0.025W and be just fine. If the circuit is outputting 51 volts at 0.1% duty cycle, now you are exceeding the voltage ratings of the part, but not the power rating.

#### MrChips

Joined Oct 2, 2009
31,079
100, A 0603 resistor , Power rating = 0.063W, and voltage rating is 50V.
So per you it is going to dissipate 25W.
Still it is not clear to me !
Thanks !
Breakdown voltage and power dissipation are maximum ratings and are not related.
You need to limit the operating conditions to both specifications. Exceed either at your own peril.
Can the resistor dissipate 25W? No. The power rating is much lower than that.
Can the resistor tolerate greater than 50V? No.

#### MisterBill2

Joined Jan 23, 2018
19,419
The solution will be to know both the current thru the resistor, the resistor value, and the resistor power rating. If the voltage that the circuit operates at is less than the voltage rating of the resistor then there is no need to be concerned about voltage, ONLY THE POWER RATING of the resistor. The POWER in the resistor, (I x I x R) is what causes heating.