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How to calculate the current across two pointsHow to calculate the current across two points in this : https://lounge.kth.se/RequestHandler...17508f458bd73b So what i did is i used Ohms law to find the total current across the whole circuit which i computed to be 3 A and i am given the EMF of the battery and asked to neglect the internal resistance. So i am stuck at calculating the current across point A-B. Please help |

Use electric induction |

It would be helpful if you could show us how you calculated the total current is 3A. What are the values of E and the four resistors? Sometimes there are "tricks" than can come into play for solving problems like this depending on the specific values of the resistors. But nevertheless, the objective is to find the current running through each resistor, because once you have that then from Kirchoff's Law the current flowing in the A-B link needs to make up for any differences between current coming into A from R1 and flowing out of A into R2, and similarly balance the current flowing into B from R3 and out of B to R4. The procedure is to start by recognizing the voltage is the same at points A and B (do you see what that is?). From this you know that the voltage drop across R1 is equal to the voltage drop across R2, and hence from Ohm's Law: $\displaystyle I_1R_1 = I_2R_2$ Similarly: $\displaystyle I_3R_3 = I_4R_4$ In addition since you know the total currents is 3 amps, from Kirchoffs Law you have: $\displaystyle I_1 + I_2 = I_3+ I_4 = 3$ Amps. That's four equations in 4 unknowns. Now you can simplify: From the first equation: $\displaystyle I_1R_1 = (3-I_1)R_2$ Solve for $\displaystyle I_1$, and then use that value to determine $\displaystyle I_2 = 3-I_1$. Similarly: $\displaystyle I_3R_3 = (3-I_3)R_4$ So now you have all four currents. Use Kirchoff to determine the flow from B to A (remember that sum of currents flowing into A must equal currents flowing out of A). Hope this helps. |

This looks like the exact same question: http://physicshelpforum.com/kinemati...ease-help.html |

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