Four resistors of 12, 3.0, 5.0, and 4.0 Ω are connected in parallel. A 12-V battery is connected to the combination. What is the current thr

Question

Four resistors of 12, 3.0, 5.0, and 4.0 Ω are connected in parallel. A 12-V battery is connected to the combination. What is the current through the battery?

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Hưng Khoa 5 years 2021-08-31T10:31:47+00:00 1 Answers 38 views 0

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  1. yenoanh
    0
    2021-08-31T10:32:50+00:00
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    Answer:

    Therefore,

    The current through the battery is 10.4 Ampere.

    Explanation:

    Given:

    V = 12 V  Battery

    Connection is Parallel,

    R₁ = 12 Ω

    R₂ = 3 Ω

    R₃ = 5 Ω

    R₄ = 4 Ω

    Let I₁, I₂, I₃, I₄ be the current passing through R₁ ,R₂, R₃, R₄  resistors

    To Find:

    I =? (current through the battery)

    Solution:

    As Connection is Parallel Voltage Remain SAME through resistors

    Bu Ohm’s Law we have

    I =\dfrac{V}{R}

    So current through R₁

    I_{1}=\dfrac{V}{R_{1}}

    Substituting the values we get

    I_{1}=\dfrac{12}{12}=1\ A

    Similarly, for current through R₂, R₃, R₄,

    I_{2}=\dfrac{V}{R_{2}}=\dfrac{12}{3}=4\ A

    I_{3}=\dfrac{V}{R_{3}}=\dfrac{12}{5}=2.4\ A

    I_{4}=\dfrac{V}{R_{4}}=\dfrac{12}{4}=3\ A

    Now in Parallel Connection we have,

    I=I_{1}+I_{2}+I_{3}+I_{4}

    Substituting the values we get

    I =1+4+2.4+3=10.4\ Ampere

    Therefore,

    The current through the battery is 10.4 Ampere.

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