When two resistors are wired in series with a 12 V battery, the current through the battery is 0.30 A. When they are wired in parallel with

Question

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.30 A. When they are wired in parallel with the same battery, the current is 1.6 A. What are the values of the two resistors?

Note: I understand other people have asked the same concept with different numbers, and their questions have been answered, however I do not understand why you need to multiple R1 by R2 when the resistors are in parallel, and furthermore, why are you supposed to multiply the “in parallel” equation by (R1+R2; the in series resistors) to find R1*R2?

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Ngọc Hoa 4 years 2021-09-02T16:55:43+00:00 1 Answers 9 views 0

Answers ( )

  1. Amity
    0
    2021-09-02T16:57:03+00:00
    Reply

    Answer:

    30\Omega, 10\Omega

    Explanation:

    Let two resistors R1 and R2 are wired in series.

    Potential difference, V=12 V

    Current=I=0.3 A

    We have to find the value of two resistors.

    When two resistors are connected in series

    R=R_1+R_2

    V=IR=I(R_1+R_2)

    Substitute the values

    12=0.3(R_1+R_2)

    R_1+R_2=\frac{12}{0.3}=40

    R_1+R_2=40..(1)

    In parallel

    \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

    \frac{1}{R}=\frac{R_2+R_1}{R_1R_2}

    R=\frac{R_1R_2}{R_1+R_2}

    Current in parallel, I=1.6 A

    V=IR

    V=1.6(\frac{R_1R_2}{R_1+R_2})

    \frac{12}{1.6}=\frac{R_1R_2}{40}

    R_1R_2=\frac{12\times 40}{1.6}=300

    R_1-R_2=\sqrt{(R_1+R_2)^2-4R_1R_2}

    R_1-R_2=\sqrt{(40)^2-4(300)}=20….(2)

    Adding equation (1) and (2)

    2R_1=60

    R_1=\frac{60}{2}=30\Omega

    Substitute the value in equation (1)

    30+R_2=40

    R_2=40-30=10\Omega

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