Razavi Basic Circuits Lec 5: More on KVL/KCL; Node Analysis

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  • Опубликовано: 28 ноя 2024

Комментарии • 7

  • @markmarkiewicz9848
    @markmarkiewicz9848 Месяц назад

    35:00:
    Everything I know, I learned from the "Great B". But, I'm still learning, and this drove me nuts for a few days. Please let me know if making K should have been "negative" on left side of = sign.
    It is the only way it works out for me.
    I tried the math and many simulations to prove it out.
    Is False:
    Ix+KI1 = I1+(I1R1/R2)+
    (I1R1/R3+R4||R5)+Iy
    Is True:
    Ix-KI1 = I1+(I1R1/R2)+
    (I1R1/R3+R4||R5)+Iy
    Also True:
    Ix = I1+(I1R1/R2)+
    (I1R1/R3+R4||R5)-KI1+Iy
    =>
    Ix-KI1 = I1+(I1R1/R2)+(I1R1/R3+R4||R5)+Iy
    Finally "I1"
    I1 = Ix-Iy/(I+(R1/R2)+
    (R1/(R3+R4||R5))+K)
    Thank You for any help!
    MGM

  • @user-in5ix1xf8x
    @user-in5ix1xf8x 8 месяцев назад

    Thank you!

  • @markmarkiewicz9848
    @markmarkiewicz9848 Месяц назад

    35:00 NO HELP YET!!!!
    What he says about current going in equals current going out makes sense. But the dependent current source other than 1 to 1 (no gain) will not follow KCL.
    The extra current has to come from somewhere.
    The formula will only work if K is reversed logic.
    Example, K being positive with the other drains
    (I1+I1R1/R2+KI1,,,etc,,)
    or negative when moved left of
    the equals sign.
    PLEASE, if anyone else has tried this example, let me know what worked out for you!
    Thank You
    MGM

  • @djura6303
    @djura6303 2 года назад

    21:00 Why doesn't he put a ground on the schematics. When inserted into the circuit simulator, the simulation without gnd does not work, and when gnd is inserted, the current through the R1 and R2 branches is 0? Thank you.

  • @kadarikruthik1241
    @kadarikruthik1241 10 месяцев назад

    11:43 how the current from voltage source and resistor are the same. as the voltage source do not follow the ohm's law.

    • @yan70172
      @yan70172 8 месяцев назад +2

      pick any point between the voltage source and the resistor, apply KCL, you can see that the current is the same.