Understanding Ohm's Law and the Relationship between Resistance and Current

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The discussion focuses on proving the relationship between ohms, joules, coulombs, and amperes in the context of Ohm's Law. Participants clarify that 1 ohm can be expressed as 1 V/A, which further breaks down to 1 m²·kg·s⁻³·A⁻². The conversion of coulombs to amperes is highlighted, noting that 1 A equals 1 C/s. Confusion arises regarding the units in the equation, particularly the presence of s⁻³ and A⁻² in the expression for ohms. Ultimately, the discussion leads to a clearer understanding of the relationships between these units in electrical contexts.
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I need to prove how 1 ohm is equal to
1 J/C*A
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J being Joules
C being coulombs
A being Amperes
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So far I got this, tell me if I'm going in the right track.
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1 J/C*A -- 1 kg*m/s^2 over C and A
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Now how can I do the rest knowing that you can find an ohm by the equation R = p*L/A? I know that a coulomb is the inverse of 6.24*10^18 and its 1 A*s.
 
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Okay so i know 1 Ω = 1 V/A = 1 m2·kg·s–3·A–2
so...
1 kg*m/s^2 over CA

C = A*S

But why is 1 Ω = 1 V/A = 1 m2·kg·s–3·A–2 has "...s–3·A–2?" That's what I'm confused about. Can someone break it down for me?
 
1 V = 1 J/C = 1 kg.m2.s-2.C-1

Since 1 A = 1 C/s,

1 V = 1 kg.m2.s-3.A-1
 
Oh! Now i see the picture! Thank you:smile:
 
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