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harina [27]
3 years ago
12

An ac generator has a maximum emf output of 150 V. What is the rms current in the circuit when the generator is connected to a 3

5 W resistor? a. 3.1 A c. 1.5 A b. 2.6 A d. 1.2 A
Physics
1 answer:
natima [27]3 years ago
3 0

This is an insidious question.  Quite frankly, I would not have
expected to see it here on Brainly.  But I'm ready to play the
cards that you have dealt me.

None of the choices offered is a correct solution.

If the output of the AC generator is nice and sinusoidal, and
its maximum (peak) emf is 150 volts, then its RMS emf is

                             (1/2) (150) (√2)  =  106.07 volts.

The resistor's dissipation is   

                             Power = (current) x (voltage) .

If the resistor is dissipating its full rated 35W, then

                       35W  =  (current) x (106.07 V)

Divide each side by 106.07 V:   

           RMS Current =  (35W) / (106.07 V)  =  0.33 Ampere .
_________________________________________

Looking over the choices offered . . . 

The largest choice ... 3.1 A ... is the current in a resistor
that is dissipating 35W if the voltage is

                           (35W / 3.1A)  =  11.29 volts .

The smallest choice ... 1.2 A ... is the current in a resistor
that is dissipating 35W if the voltage is

                           (35W / 1.2A)  =  29.17 volts .

Whatever you meant the so-called "150 V" of the generator
to represent ... whether the RMS sinusoidal, peak sinusoidal,
peak square-wave, RMS square-wave, DC, average, etc. ...
none of the choices for current, in combination with any of these
generators, would dissipate 35W.

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lutik1710 [3]

a) E = 0

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Explanation:

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We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

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Here we want to find the electric field at a distance of

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Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

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E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

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Answer:

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At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

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Explanation:

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Answer:

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6 0
3 years ago
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