The ideal ammeter in the given circuit shown reads K Ampere. Then what is the value of K?

butthurtnevergetsbetter

Wasted 5 resistors in that shoddy design.

8 years ago |

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IamAmanWhoWalksAloneAndWhenImWalkingAdarkRoad

Do your own homework you lazy fuck.

8 years ago |

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StarscreamAndHutch

3.14

8 years ago |

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wadenelson1

10v / 3.33 ohms = 3.33 A. You've got 3 ten amp circuits in parallel = 3.33 ohms. the added resister is between equipotential points 1/2

8 years ago |

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wadenelson1

Sorry, three ten OHM circuits in parallel.

8 years ago |

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quetzalcoatl360

10 / 3.33 = 3 !?

8 years ago |

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voidfury

Just to be clear, you're using a symmetry argument to say the current on that middle wire is zero. (Flipping those two branches=no change)

8 years ago |

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wadenelson1

Correct. It's a visual distraction, but if current through a branch is zero, so is the voltage drop through that branch.

8 years ago |

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voidfury

If you don't have such a nice way to simplify the problem, you can use Kirkoff's loop and junction laws to solve this type of problem.

8 years ago |

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