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Jamesman174

According to Ohm's law what would happen to the current if resistance was increased? Increase Decrease varys nothing

  • 2 years ago
  • 2 years ago

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  1. Schrodinger
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    This sounds a LOT...like a test question...mm? What is this for?

    • 2 years ago
  2. Jamesman174
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    My friend is texting me asking me and I have no idea :P He helps me with math and normally science is my strong point XD. I think it would be physical science.

    • 2 years ago
  3. Schrodinger
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    I = V/R, correct? Then what would happen to I if R was increased?

    • 2 years ago
  4. Jamesman174
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    I think it would be for his physical science test or something* Didn't put that last bit in :P. I have no idea, I have never heard of this stuff before :P

    • 2 years ago
  5. sylvstrknnth
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    R decrease as we keep V as a constant.. if im not mistaken

    • 2 years ago
  6. Schrodinger
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    Well, if you can't make an assumption off of that, you're helpless. It's basic math. I = V/R. What happens to I if R increases. I don't want to just give you the answer, because that's cheating. It's not rocket science, man, and I can't trust if this is you asking the question or if it's your friend. Either way, use your head.

    • 2 years ago
  7. Kainui
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    Since I=current, V=voltage, and R=resistance and from Ohm's law we can clearly see I=V/R then let's consider what we know about how fractions work. As the numerator (top) increases, it makes a number bigger. Let's think about x/1. So if you say X=1, it's 1/1=1. X=2, it's 2/1=2. Makes sense? As the denominator (bottom) increases, it makes a number smaller. 1/x, if we plug in 1 you get 1/1=1. For 2 you get 1/2 which is a half. It makes sense that as the bottom increases, the number gets smaller. I hope that helps build some intuition on how fractions work so you understand and can think, alright, so for the fraction V/R does the fraction get bigger as the denominator increases for a constant numerator?

    • 2 years ago
  8. Jamesman174
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    Well I told him everything you all said and I guess he got it right, so thanks for your help.

    • 2 years ago
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