Problem: In the Circuit shown, notices that the values of some currents are given, so that instead of unknown currents,

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answerhappygod
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Problem: In the Circuit shown, notices that the values of some currents are given, so that instead of unknown currents,

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Problem In The Circuit Shown Notices That The Values Of Some Currents Are Given So That Instead Of Unknown Currents 1
Problem In The Circuit Shown Notices That The Values Of Some Currents Are Given So That Instead Of Unknown Currents 1 (266.2 KiB) Viewed 121 times
Problem: In the Circuit shown, notices that the values of some currents are given, so that instead of unknown currents, the unknowns are the emf's ξ1​ and ξ2​, and the value of the resistor R, and the current through the 3Ω resistor. Four unknowns need four equations for a solution. Carefully write down the Kirchoff circuit laws for this problem and show your work as you solve for the unknown values. Input your answers into the Online Homework Ch24_Ex_6_Kirchoff quiz by tomorrow and 10:00 AM and turn in your work for this problem. (Hint: I used three small loops and the node beneath the 3 Ohm resistor.
Problem 2: The capacitor in the figure is initially uncharged. At t=0, a switch (not shown) is closed connecting the battery ξ through the rest of the circuit through R1​. Find the time at which the voltage across the capacitor reaches half of its maximum value. Turn in your solution with your work. Input your answer online. (Hints: Use two simple Kirchoff loops and one node for the currents going through R1, R2 and R3, remembering that I3​=dtdq​ the current into the capacitor, and that the voltage drop across the capacitor at any time is CQ(t)​. You will have three equations in the unknowns, I1​,I2​ and I3​. Use ordinary algebra and two the equations to solve for and eliminate I1​ and I2​ symbolically, leaving only a single equation involving q and I3​. Put this in the form of the equation dtdQ​+RC1​Q=ξ and read off the RC time, and the final voltage across the capacitor. Then use the equation V(t)=ξ(1−e−t/RC) to find the time the voltage is equal half its maximum value. ... I find that the maximum value is 18 volts.)
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