A 16 Ω resistor and a 6 Ω resistor are connected in series to an ideal 6 V battery. Find the current in each resistor. Answer in

Question

A 16 Ω resistor and a 6 Ω resistor are connected in series to an ideal 6 V battery.
Find the current in each resistor.
Answer in units of A.
b) Find the potential difference across the first
resistor.
Answer in units of V.
c) Find the potential difference across the second
resistor.
Answer in units of V

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Thanh Thu 3 years 2021-08-27T04:30:13+00:00 2 Answers 16 views 0

Answers ( )

  1. Bo
    0
    2021-08-27T04:31:49+00:00
    Reply

    Answer:

    Explanation:

    a) series resistors carry the same current

    A = V/Re = 6/(16 + 6) = 0.2727272… ≈ 27 mA

    b) V = V₀(R/Re) = 6(16/(16 + 6)) = 4.363636 ≈ 4.4 V

    c) V = V₀(R/Re) = 6(6/(16 + 6)) = 1.636363 ≈ 1.6 V

     or V = 6 – 4.4 = 1.6 V

  2. bonexptip
    0
    2021-08-27T04:32:01+00:00
    Reply

    Answer:

    (a)  0.273 A

    (b) 4.368 V

    (c) 1.638 V

    Explanation:

    From the question,

    (a) Applying ohm’s law

    V = IR’…………………. Equation 1

    Where V = Voltage of the battery, I = Current in each of the resistor, R’ = Total resistance of the combined resistors

    Since the Two resisstors are connected in series,

    (i) The same current flows through both resistors

    (ii) The total resistor (R’) = R₁+R₂

    Therefore,

    V = (R₁+R₂)I

    Make I the subject of the equation

    I = V/(R₁+R₂)…………….. Equation 2

    Given: V = 6 V, R₁ = 16 Ω, R₂ = 6 Ω

    Substitute into equation 2

    I = 6/(16+6)

    I = 6/22

    I = 0.273 A

    (b) The potential difference across the first resisto

    V₁ = IR₁…………………. Equation 3

    Given: I = 0.273 A, R₁ = 16 Ω

    Substitute these values into equation 3

    V₁ = 0.273(16)

    V₁ = 4.368 V

    (c) The Potential difference across the second resistor is

    V₂ = IR₂……………….. Equation 4

    V₂ = 0.273×6

    V₂ = 1.638 V

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