Question

9. [03.03]

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

Answer 1

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

Answer 2

Circuit one will have more current than circuit two

Step-by-step 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