
09272005, 07:39 PM #1
Anyone who knows electrical formula's, please help
Hey, under Ohm's law, subsection Power I am trying to figure out a question.. Any help would be greatly appreciated.
"A 1/2 watt resistor has a resistance of 1000 ohms. What is hte maximum current it can handle?

09272005, 07:43 PM #2
V = I R.........sorry, that's all i remember....i hate physics

09272005, 07:44 PM #3
I know the answer.
But since you wanted to go to Queens.
Do your own homework!

09272005, 08:00 PM #4
1/2 watt resistor? watts are J/s and resistors are measured in ohm meters. max current would be potential difference over resistance. I=V/R and R = p l/a which is the resistivity of the metal times the length over area. resistivity varies depending on what metal it is (provided it is ohmic). Is there anymore to the question? It's been awhile since I've done this stuff.

09272005, 08:06 PM #5
Ignore my last post I'm a little slow when I've been drinkin. so you know power is equal to the current squared times the resistance. essentailly P=I^2 R so I = sqrt(P/R) = sqrt(.5watt/1000 ohm) = .022 amps.
this could be wrong I don't know like I said it's been long time.
edit: I think i am wrong, just doesn't seem right. sorry manLast edited by symatech; 09272005 at 08:54 PM.

09272005, 09:11 PM #6
wrong one just a minute

09272005, 09:16 PM #7
The voltage divider rule can be used with resistive, inductive, or capacitive circuit elements. It can also be used with AC or DC input sources. The equation for calculating the output voltage is different, however, depending on the type of circuit element. Following are the three general cases of two like elements connected in series:
Resistive divider:
The formula for determining the DC or AC output voltage of a resistive divider is:
Vout = Vin*R2/(R1+R2)
Example: In the following circuit, the output voltage would be: Vout = 9V*10K/(10K + 5K) = 6V
Inductive divider:
Inductive dividers can be used with AC input signals. A DC input voltage would split according to the relative resistances of the two inductors by using the resistive divider formula above. The formula for determining the AC output voltage of an inductive divider (provided the inductors are separate, i.e. not wound on the same core, and have no mutual inductance) is:
Vout = Vin*L2/(L1+L2)
Example: In the following circuit, the output voltage would be: Vout = 10VAC*50mH/(50mH + 100mH) = 3.33VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the inductors is not high at the frequency of operation (i.e. inductance not large enough), there will be a very large current drawn by the shunt element.
Capacitive divider:
Capacitive dividers can be used with AC input signals. A DC input voltage would not pass through the capacitors, so the DC case is not relevant. The formula for determining the AC output voltage of a capacitive divider is different from the resistive and inductive dividers, because the series element, C1 is in the numerator instead of the shunt element, as shown below:
Vout = Vin*C1/(C1+C2)
Example: In the following circuit, the output voltage would be: Vout = 10VAC*0.022uF/(0.022uF + 0.01uF) = 6.875VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the capacitors is not large at the frequency of interest (i.e. capacitance not large enough), the output current capability will be very low.

Copyright © 1999, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.
Revised 01/01/00

09272005, 09:20 PM #8
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I =AMPERES ~or~ (A = AMPERES)
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PF = POWER FACTOR
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DIRECT CURRENT
AMPS= WATTS÷VOLTS I = P ÷ E A = W ÷ V
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HORSEPOWER= (V x A x EFF)÷746
EFFICIENCY= (746 x HP)÷(V x A)
AC SINGLE PHASE ~ 1ø
AMPS= WATTS÷(VOLTS x PF) I=P÷(E x PF) A=W÷(V x PF)
WATTS= VOLTS x AMPS x PF P=E x I x PF W=V x A x PF
VOLTS= WATTS÷AMPS E=P÷I V=W÷A
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HORSEPOWER= (V x A x EFF x PF)÷746
POWERFACTOR= INPUT WATTS÷(V x A)
EFFICIENCY= (746 x HP)÷(V x A x PF)
AC THREE PHASE ~ 3ø
AMPS= WATTS÷(1.732 x VOLTS x PF) I = P÷(1.732 x E x PF)
WATTS= 1.732 x VOLTS x AMPS x PF P = 1.732 x E x I x PF
VOLTS= WATTS÷AMPS E=P÷I
VOLTAMPS= 1.732 x VOLTS x AMPS VA=1.732 x E x I
HORSEPOWER= (1.732 x V x A x EFF x PF)÷746
POWERFACTOR= INPUT WATTS÷(1.732 x V x A)
EFFICIENCY= (746 x HP)÷(1.732 x V x A x PF)
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09272005, 11:28 PM #9Originally Posted by needmorestrength
P=IE (Power, in Watts, = current times the voltage)
E=IR (Voltage = current time the resistance)
So, substitute (IR) for the E in P=IE, and you get
P=I(IR)
If P=.5 watts
If R=1000
then .5=I(I*1000)
divide both sides by 1000, and you get
.5
 = I squared
1000
or,
1
 = I squared
2000
or
.0005 = I squared
take the square root of both sides, and you get
.022 = I
So, the current would be .022 amps, or 22 milliamps.
What do I win? It's been 25 years since I screwed with this stuff . . .
If you have .022 amps flowing through a 1000 ohm resistor, what would the voltage be across it?
Well, E=IR,
so you'd multiply .022 by 1000, and you'd get 22 volts.
Tock

09272005, 11:30 PM #10
hey that's what I gots!! except he's not looking for voltage he's looking for current. .022 amps. I hope you're right b/c that would mean I was right and because of whiskey and beer I like being right.
*sigh* i think it's time for bed, im almost drunk

09272005, 11:36 PM #11
no no NO!!! scratch that!.. I am drunk

09282005, 08:21 AM #12
....As Tock and Sym stated P=I(IR)..
so, I=√P/R
Current = √.5w/1kΩ =.o223A= 22.3mAThere are 3 loves in my life: my wife, my English mastiffs, and my weightlifting....Man, my wife gets really pissed when I get the 3 confused...
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