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