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Consider a circuit with a 1 $\Omega$ resistor and a 2 $\Omega$ resistor in parallel. If it takes twice as much energy to send an electron through a 2 $\Omega$ resistor than a 1 $\Omega$ resistor, why do any electrons flow through the 2 $\Omega$ resistor? Wouldn't it be more efficient to send all the electrons through the 1 $\Omega$ resistor?

Current flows through both branches of the parallel circuit, so this logic is obviously flawed, but I don't understand why.

Shamina
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Rational Function
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  • Maybe they need teamwork to get through the resistance :D – Emil Jan 06 '17 at 12:01
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    Current flows because there's a potential applied, and Ohm's law describes current as a result. The 'more efficient' concept is not applicable. – Whit3rd Jan 06 '17 at 13:10
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    The question is essentially equivalent to "Why does water flows through a small hole in a barrel if there is a bigger one in it" – user_na Jan 06 '17 at 13:22
  • The battery does not have a mind. It does not decide which route is more efficient. – sammy gerbil Jan 06 '17 at 14:19

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It doesn't take twice as much energy. Both resistors in parallel have the same voltage drop so it take the same amount of energy for any given charge to cross either path. The path with lower resistance dissipates more energy per time exactly because there are more charges flowing past it per time.

Consider a bathtub with a large and a small hole falling to the same elevation. Any given drop of water loses the same potential energy no matter which hole it falls, but the larger hole can accommodate a greater flow rate.

octonion
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    Thanks @octonion. I find your analogy really helpful for understanding this concept. It also didn't occur to me that "The path with lower resistance dissipates more energy per time exactly because there are more charges flowing past it per time." Your answer explains the idea very well. – Rational Function Jan 07 '17 at 15:30