What is the maximum voltage that can be applied to a 2.7 k resistor rated at ¼ watt?
2007-12-10 20:16:29 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Wattage is E^2/R, so E = √RW
E = √[2.7*10^3*0.25] = 26v
2007-12-10 20:30:38 · answer #1 · answered by gp4rts 7 · 2⤊ 2⤋
Please bear with me, I'm a mechanical engineer... this may not be 100% correct, but I'll give it a shot:
Since V=i*R and P=V*i (i=P/V)
we can substitute (i=P/V) into V=i*R
and get V=P/V*R which is V²=P*R or V=â(P*R)
Where V is Voltage, P is Power (watts), R is resistance (ohms), and i is current (amps)
Voltage is then equal to the square root of the product of power (¼ watt) and resistance (2,700 ohms) or 25.98 Volts.
The maximum voltage that can safely be applied will be less than 25.98 Volts if the resistor is rated at ¼ Watt. I don't know what an appropriate safety factor would be for this, but just be aware that anything above 25.98V will overload the resistor.
2007-12-10 20:35:02 · answer #2 · answered by Anonymous · 1⤊ 1⤋
Power = (Voltage)^2 / Resistance
1/4 = V^2 / 2700
V^2 = 675
V = 25.98 volts.
2007-12-10 22:55:58 · answer #3 · answered by gauravragtah 4 · 0⤊ 1⤋
i may have used the wrong fomula but....
25.9 Volts.
V=square root of PxR
2007-12-10 20:32:40 · answer #4 · answered by flashlight 2 · 1⤊ 2⤋