Question

6. For a half wave rectifier average value of output current is (io is the maximum value of current)
(A) i0
(B) i0π
(C) 2i0π
(D) 2/i0π​

Answer 1

Average current in HALF WAVE RECTIFIER:

$1) \: i = i_{0} \sin( \omega t) \: \: \: \: . \: . \: . \: .(t < \dfrac{T}{2} )$

$2) \: i = 0 \: \: \: \: . \: . \: . \: .( \dfrac{T}{2} < t < T )$

Now, average current will be :

$\therefore \: i_{avg} = \dfrac{ \displaystyle\int_{0}^{T} i \: dt }{ \displaystyle\int_{0}^{T} \: dt}$

$\implies \: i_{avg} = \dfrac{ \displaystyle\int_{0}^{T} i \: dt }{T}$

$\implies \: i_{avg} = \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} } i_{0} \sin( \omega t) \: dt + \int_{ \frac{T}{2}}^{T} 0 \: dt}{T}$

$\implies \: i_{avg} = \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} } i_{0} \sin( \omega t) \: dt }{T}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} } \sin( \omega t) \: dt }{T}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \bigg \{ - \cos( \omega t) \bigg \}_{0}^{ \frac{T}{2}}}{ \omega T}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \bigg \{ - \cos( \omega t) \bigg \}_{0}^{ \frac{T}{2}}}{ \frac{2\pi}{T} \times T}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \bigg \{ - \cos( \omega t) \bigg \}_{0}^{ \frac{T}{2}}}{2\pi}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \bigg \{ \cos( 0) - \cos( \frac{2\pi}{T} \times \frac{T}{2} ) \bigg \}}{2\pi}$

$\implies \: i_{avg} = i_{0} \times \dfrac{ \bigg \{1 - ( - 1) \bigg \}}{2\pi}$

$\implies \: i_{avg} = i_{0} \times \dfrac{2}{2\pi}$

$\implies \: i_{avg} = \dfrac{i_{0}}{\pi}$

So, average value of current is i_(0)/π.