Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel are linked in series to a 3.0 Ω resistor. All three of this is in parallel

Question

Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel are linked in series to a 3.0 Ω resistor. All three of this is in parallel with a fourth resistor. If the total effective resistance is 1.0 Ω, what is the resistance of the fourth resistor?
a)11/3
b)11/8
c)4
d)3

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Adela 4 years 2021-08-30T12:53:20+00:00 1 Answers 12 views 0

Answers ( )

  1. mit
    0
    2021-08-30T12:54:40+00:00
    Reply

    Answer: 11/8

    Explanation:

    Given the following:

    R1 = 1.0 Ω, R2 = 2.0 Ω, R3 = 3.0 Ω

    Total effective resistance = 1.0 Ω

    Resistance in parallel :

    R = (R1×R2) / R1+R2

    Resistance in series :

    R = R1 + R2 + …. +Rn

    Therefore,

    R1 and R2 in parallel

    Re = (1 × 2) / (1 + 2) = 2/3 Ω

    Re is linked in series to R3

    Rf = Re + R3

    Rf = 2/3 + 3 = 11/3

    Rf linked in parallel to R4 with effective resistance being 1 Ω

    11/ 3 and R4 in parallel gives 1 Ω

    Re = (11/3 × R4) / (11/3 + R4)

    1 = 11/3 R4 / 11/3 + R4

    Cross multiply

    11/3 + R4 = 11/3 × R4

    11/3 = 11/3R4 – R4

    11/3 = 8/3R4

    Multiply both sides by 3/8

    33/24 = R4

    IN LOWEST TERM

    R4 = 11/8

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