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Study asked in Science & MathematicsPhysics · 2 months ago

In the circuit shown in the figure, R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 20.0 Ω, and Vemf = 12.0 V. will give fave ans physics?

In the circuit shown in the figure, R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 20.0 Ω, and Vemf = 12.0 V.

a. Determine a value for the equivalent resistance.

b. Calculate the magnitude of the current flowing through R3 on the top branch of the circuit (marked with a vertical arrow). 

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2 Answers

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  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    This is made a heck of a lot easier due to the symmetry of the top and bottom. For the R1/R2 in parallel,

    R12 = 1/(1/3.00 + 1/6.00) = 1/(3/6.00) = 2.00 Ω

    and so for the top and bottom halves of the circuit

    Rt = Rb = 2*2.00Ω + 20.0Ω = 24.0Ω

    Applying Kirchoff's Law on the top loop:

    E = 12.0 V = It * Rt = It * 24.0Ω → → It = 0.500 A ◄ (b)

    and of course for the bottom we get Ib = 0.500 A

    (a) Not sure about this part. I'd wager that

    Req = 12.0V / 0.500A = 24.0 Ω

    but it's possible that we should use the total current, and so

    Req = 12.0V / 2*0.500A = 12.0 Ω

    Attachment image
  • 2 months ago

    b.

      3//6=2Ω

    2+20+2+=24Ω

    Rt=24//24=12Ω

    a.

    It=12/12=1A

    I(3Ω)=1/2A

     

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