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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:

The process involved in this problem are:

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Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

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Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

For process 2:

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$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

For process 3:

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

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Answer:

The temperature of the windings are 60.61 °C

Explanation:

Step 1: Data given

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Temperature = 20.0 °C

After the motor has run for several hours the resistance rises to 58Ω.

Step 2: Calculate the new temperature

Formula: R = Rref(1 + α(T-Tref))

⇒with α = temperature coëfficiënt of Cupper at 20 °C = 0.00394/°C

⇒with Tref = reference temperature = 20°C

⇒with T = end temperature = TO BE DETERMINED

⇒with R = resistance at end temperature = 58Ω

⇒with Rref = resistance at reference temperature = 50 Ω

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4 years ago