In need of assistance.

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codan

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Hi Everybody,

I decided to do an electronics course but i am confused with a question they ask.

They say:
A 4700 ohm resistor is rated at .5 Watts. What is the maximum safe working current.

This i am having trouble with as they do not specify a Voltage.

If i use Ohms Law i have

I = The Square Root of P/R

I = Amperes
P = Watts
R= Resistance

But how can i answer the question if i do not know the Voltage value.

If
P = V x I then the Amperage value will change due to the Voltage value.

Eg
1v X .5A = .5Watts
10v X .05A = .5Watts

Am i missing the obvious here?

Thank You
 

hi,

Watts= I^2 * R

So: I^2 = W/R

I= √ (W/R)
 
Hi Eric,

Thanks for the reply, i guess i was thinking about it the wrong way.
I'm not sure i fully understand why voltage is not included in the calculations though.

Thank you
 
A Google image search for Ohm's law gives this very helpful graphic.

I keep it handy to save me the trouble of having to work out how to calculate current and voltage from power and resistance.
**broken link removed**
 
Actually, you could say that volts are in the equation. Since V=IR and P = I^2 R you could substitute voltage back in to the power equation getting P=VI
 
codan said:
I'm not sure i fully understand why voltage is not included in the calculations though.
Click to expand...

Consider this:

W = V²/R so V²= WR

I = V/R = (√W.R)/R

I = √W/√R = √(W/R)

Which is exactly what Eric arrived at by a different route.

JimB
 
cowana said:
Voltage was in the equation - you just replaced V with IR.

Andrew
Click to expand...

hi,
There is only one voltage value that would drive current thru the 4700R to make it dissipate 0.5Watts.!

V^2= WR so V=√(W*R)

V^2 = 0.5 *4700

So V= 48.5v
 
Last edited:
All Ohm's law shows you the maximum continuous voltage and current .

In reality, the resistor will be able to take higher voltages and currents for a short period of time.

The peak maximum voltage rating will be specified on the datasheet.
 
Thanks to everyone for the replies & help it is greatly appreciated, i now have a better understanding.
Eric's last reply made it a lot easier to understand.

Thanks again
 
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