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4 years ago

9. [03.03]

Physics

2 answers:

Evgen [1.6K]4 years ago

5 0

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

gizmo_the_mogwai [7]4 years ago

3 0

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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Explanation:

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$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

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$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

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$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

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