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Nuetrik [128]
3 years ago
13

wo different electrical devices have the same power consumption, but one (device 1) is meant to be operated on 127 V AC and the

other (device 2) on 230 V AC. (a) What is the ratio of their resistances? R2 R1 = (b) What is the ratio of their currents? I2 I1 = (c) Assuming its resistance is unaffected, by what factor will the power increase if a 127 V AC device (device 1) is connected to 230 V AC?
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

a) R₂/R₁ = 0.3049

b) I₂ / I₁  =  1.81

c) (230)² / ( 127)² =  3.28

Explanation:

By definition  the power consumption is:

P = I²*R    ⇒  as  I = V/R   I² = (V/R)²    and  P = (V²/R²)*R

P = V²/R

P = power consumtion  

I = current

R = is the resistor

Case Device 1   Voltage 127 (V)

P₁  = V₁² /R₁     P₁  =  (127)²/ R₁

Case Device 2 Voltage  230 (V)

P₂  = ( 230)²/ R₂

As   P₁  =  P₂        (127)²/R₁  =  (230)²/R₂

(127)²/(230)²  =  R₂/R₁

a) R₂/R₁ = 0.3049

b)  P₁ = I₁²*R₁              P₂  =  I₂²*R₂

I₂² *R₂  =  I₁²*R₁

I₂² / I₁²   =  R₁/R₂

I₂² / I₁²   =  1/ 0.3049

I₂² / I₁²   =  3.2797

√  (I₂² / I₁² )   =  √3.2797

b) I₂ / I₁  =  1.81

c)  P₁  = V₁*I₁      ⇒   P₁ = V₁²/R₁      if we "use" the device in 230 (V)

P₁´  =  V₂² / R₁

P₁  =  (127)²/R₁      and     P₁´ = (230)²/ R₁

Then the increasing factor is:

c) (230)² / ( 127)² =  3.28

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Explanation:

Given the angular frequency (\omega) of the simple harmonic oscillator is doubled.

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a_{max}=A\omega^2

Now, according to the problem, the angular frequency (\omega) got doubled.

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a_{max'}=4a_{max}

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

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A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
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 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
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Hence, The percentage error in his experimental value is -51.97%.

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Explanation:

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Solution

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