To find the average output voltage ((V_{avg})) for a type D chopper, we can use the following formula:

[

V_{avg} = V_{supply} times frac{t_{on}}{T}

]

where:

– (V_{supply}) is the supply voltage,

– (t_{on}) is the time period the switches are on,

– (T) is the total period of the chopper frequency.

Given:

– (V_{supply} = 220 , V)

– Chopper frequency (f = 2 , kHz = 2000 , Hz)

– (t_{on} = 0.3 , ms = 0.0003 , s)

First, we calculate the total period (T):

[

T = frac{1}{f} = frac{1}{2000} = 0.0005 , s = 0.5 , ms

]

Now we can calculate the average output voltage:

[

V_{avg} = 220 times frac{0.0003}{0.0005}

]

Calculating the fraction:

[

frac{0.0003}{0.0005} = 0.6

]

Now substituting back:

[

V_{avg} = 220 times 0.6 = 132

To solve this, we first need to find the value of ( T ), the period of the chopping frequency.

The chopping frequency ( f ) is given as 2 kHz, which is equal to ( 2000 ) Hz.

The period ( T ) can be calculated using the formula:

[

T = frac{1}{f}

]

Plugging in the value for frequency:

[

T = frac{1}{2000} = 0.0005 text{ seconds} = 0.5 text{ ms}

]

Next, we know that ( T_{text{on}} = frac{1}{4} T ). Therefore, we can calculate ( T_{text{on}} ):

[

T_{text{on}} = frac{1}{4} times 0.0005 = 0.000125 text{ seconds} = 0.125 text{ ms}

]

Now, in a constant frequency system, the total time ( T ) is equal to the sum of ( T_{text{on}} ) and ( T_{text{off}} ):

[

T = T_{text{on}} + T_{text{off}}

]

Rearranging gives us:

[

T_{text{off}} = T – T_{text{on}}

]

Now we substitute the values we have:

[

T_{text{