9. [03.03] When comparing two circuits, you note that circuit one has twice the voltage and half the resistance of circuit two.​

Question

9. [03.03]
When comparing two circuits, you note that circuit one has twice the voltage and half the resistance of circuit two.​

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Maris 2 months 2021-08-21T22:50:48+00:00 2 Answers 0 views 0

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    0
    2021-08-21T22:52:31+00:00

    Answer:

    Circuit one will have more current than circuit two

    Explanation:

    I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm’s Law, which states the following –

    V = IR,

    where V = voltage / potential difference, I = current, and R = resistance

    If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance –

    2V = I( 1 / 2R ),

    4V = IR,

    I = 4V / R

    Whereas in the second circuit –

    V = IR,

    I = V / R

    As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

    Hence, circuit one will have more current than circuit two

    0
    2021-08-21T22:52:40+00:00

    That’s a very interesting observation.  When you notice that, does it stimulate any questions in your mind due to your natural curiosity ?

    For example, do you walk away wondering, perhaps, how the current in the two circuits might compare ?  If so, you’ve come to the right place.  Read on:

    Take the form of Ohm’s Law for current . . . I = V / R . . . and apply it to both circuits:

    For the second circuit:  I₂ = V / R

    For the first circuit:  

    I₁ = (2V) / (R/2)

    Multiply numerator and denominator by 2 :

    I₁ = (4V) / R

    I₁ = 4 (V / R)

    I₁ = 4 I₂  

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