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3 years ago

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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

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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Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

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Solution:-

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$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

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solong [7]

Answer:

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