Respuesta :
Answer:
V_{average} = [tex]\frac{1}{2} V_o[/tex] , 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) = [tex]\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.[/tex]
to substitute in the equation let us separate the into two pairs
V_average = [tex]\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt[/tex]
V_average = [tex]V_o \ \int\limits^{1/2}_0 {} \, dt[/tex]
V_{average} = [tex]\frac{1}{2} V_o[/tex]
we evaluate V₀ = 4 V
V_{average} = 4 / 2)
V_{average} = 2 V
