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azamat
3 years ago
14

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

ver one period of the waveform. Hint: the equation for a square wave will be a piecewise function and it will be convenient to start the integration where the voltage changes; for example in this problem we could define that during the first half of the period the voltage is 0 V and for the second half of the period the voltage is 4 V
Physics
1 answer:
Margarita [4]3 years ago
6 0

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

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Lapatulllka [165]

The electric field between plates is 4000V/m.

An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.

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The voltage between points A and B is

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Given:

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I think this is correct, but I am not entirely certain.

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Answer:

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T₁ = 745 K

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S_1=-\dfrac{7190}{745}\ J/K

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Pachacha [2.7K]

Explanation:

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(B). Given that,

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Hence, This is the required solution.

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