You are given a number of 12 Ω resistors, each capable of dissipating only 2.6 W without being destroyed. What is the minimum number of such

Question

You are given a number of 12 Ω resistors, each capable of dissipating only 2.6 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 12 Ω resistance that is capable of dissipating at least 10.1 W?

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Helga 5 months 2021-09-05T10:18:46+00:00 1 Answers 0 views 0

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    2021-09-05T10:20:09+00:00

    Answer:

    For series,

    n = 4 resistors

    For parallel

    n = 4 resistors

    Explanation:

    From the question,

    For Series

    Note: When resistors are connected in series, the same current flows through each of the resistors and the combined resistance

    If 12Ω resistor each is capable of dissipating 2.6 W.

    P = I²R…………….. Equation 1

    Where P = power, R = resistance of the resistor, I = current.

    make I the subject of the equation

    I = √(P/R)…………. Equation 2

    Given: P = 2.6 W, R = 12 Ω

    Substitute into equation 2

    I = √(2.6/12)

    I = 0.4655 A

    The combined number of 12 Ω resistor in series is given as

    R’ = Rn…………….. Equation 3

    Where n = minimum number of resistor, R’ = combined resistance

    R’ = 12n

    10.1 = (0.4655)²12n

    10.1 = 0.2167×12n

    10.1 = 2.6n

    n = 10.1/2.6

    n = 3.88

    n ≈ 4 resistor

    For parallel,

    Note: when resistors are connected in parallel, The have the same voltage with the combined resistance

    P = V²/R………………… Equation 4

    make V the subject of the equation

    V = √(PR)

    V = √(2.6×12)

    V = √31.2 V

    The combined resistance for parallel connection

    R’ = R/n

    R’ = 12/n

    Substitute into equation 4

    10.1= (√31.2)²n/12

    10.1×12 = 31.2n

    121.2 = 31.2n

    n = 121.2/31.2

    n = 3.88

    n = 4 resistors

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