ThatOtherChigga #Mack Posted September 26, 2013 Share Posted September 26, 2013 Question #1: A 1/4 watt resistor is connected to a six volt source. How much current is required to reach the maximum 1/4 rating of the resistor. So is it 1/4*6 = C or is it a different equation? (Current is measured in Ohms right?) Question #2: A 1/2 watt 1200 ohm resistor is connected to a power source. What is the maximum voltage that can be applied to math its power rating? So what's the equation, and why? Question #3: A circuit design specification states that all carbon composite resistors used shall not exceed 50% of their rated wattage. What is the maximum current a 1.2 kiloOhm, 1/4 watt carbon composite resistor can safely carry and meet the specifications? Equation and why? Ohm's Law: E=I*R Watt's law: P=I*E Don't forget stuff like E=P/I, I=E/R, etc. Link to comment Share on other sites More sharing options...
Pyroman Posted September 26, 2013 Share Posted September 26, 2013 You're asian, you're supposed to be able to solve that in less than a nanosecond. Link to comment Share on other sites More sharing options...
Triforce of Charmanders Posted September 26, 2013 Share Posted September 26, 2013 Wait... non shit post from Chigga? Monitor must be broken >turns off and on Link to comment Share on other sites More sharing options...
Distinctively Vincent Posted September 26, 2013 Share Posted September 26, 2013 #1- Current is measured in Amps. Resistance is measured in Ohms. #2 and #3- You'll have to use both equations together. Power, voltage, current, and resistance are all connected to each other by the two laws. Link to comment Share on other sites More sharing options...
cranwell96 Posted September 26, 2013 Share Posted September 26, 2013 Here's what I got: #1 I= P/V = 0.25W/6V = 0.0417A #2 V^2= PR = 0.25W*1200ohm =0.5W*1200ohm = 300V^2 =600V^2 V= 17.3V V =24.5V #3 I^2= P/R = 0.125W / 1200ohm = 0.000104 A^2 I= 0.01A Link to comment Share on other sites More sharing options...
Triforce of Charmanders Posted September 26, 2013 Share Posted September 26, 2013 1.21 GIGAWHATTS Link to comment Share on other sites More sharing options...
There Posted September 26, 2013 Share Posted September 26, 2013 inb4 chiggawatts Link to comment Share on other sites More sharing options...
Santa Heavy Posted September 26, 2013 Share Posted September 26, 2013 Topic Name: Chigga gets backpack.tf members to do his homework for him. Link to comment Share on other sites More sharing options...
ThatOtherChigga #Mack Posted September 26, 2013 Author Share Posted September 26, 2013 Topic Name: Chigga gets backpack.tf members to do his homework for him. Eh, yeah it's my homework, but I need smart people to explain how to do the problems to me. And you guys are smart, right? <3 Link to comment Share on other sites More sharing options...
ThatOtherChigga #Mack Posted September 26, 2013 Author Share Posted September 26, 2013 Here's what I got: #1 I= P/V = 0.25W/6V = 0.0417A #2 V^2= PR = 0.25W*1200ohm = 300V^2 V= 17.3V #3 I^2= P/R = 0.125W / 1200ohm = 0.000104 A^2 I= 0.01A Someone please confirm ^^^ edit: for #2, wouldn't it be 0.5W*1200ohm? final answer ~ 24.497V Thanks crannybabe <3 Link to comment Share on other sites More sharing options...
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