To find the peak commutating current in a current commutated chopper circuit, we can use the formula derived from the relationships between the inductor (L), capacitor (C), and the applied voltage (V).

The peak commutating current (I_peak) can be calculated using the following formula:

[

I_{peak} = sqrt{frac{V^2}{frac{L}{C}}}

]

Given:

– V = 230 V

– L = 20 μH = 20 x 10^{-6} H

– C = 50 μF = 50 x 10^{-6} F

Now, substituting the values into the formula:

1. Calculate (frac{L}{C}):

[

frac{L}{C} = frac{20 times 10^{-6}}{50 times 10^{-6}} = frac{20}{50} = 0.4 text{ seconds}

]

2. Substitute into the equation for I_peak:

[

I_{peak} = sqrt{frac{230^2}{0.4}} = sqrt{frac{52900}{0.4}} = sqrt{132250} approx 363.42 text{ A}

]

Therefore, the peak commutating current is approximately 363.42 A.

To calculate the value of the capacitor (C) required for the commutating circuit in a voltage commutated chopper feeding power to an electric car, we can use the formula related to the energy storage in a capacitor and the turn-off time of the SCR.

The required formula is:

[ C = frac{I cdot t}{V} ]

Where:

– ( I ) = load current (in Amperes)

– ( t ) = turn-off time of the SCR (in seconds)

– ( V ) = voltage across the capacitor (in Volts)

Given:

– Battery voltage ( V = 60 , V )

– SCR turn-off time ( t = 20 , mu s = 20 times 10^{-6} , s )

However, to complete this calculation, we need the load current ( I ). If you provide me with that value, I can help you find the required capacitance ( C ).

Without the value of load current, I cannot provide a numerical answer. Thus, please provide the load current for a complete calculation. If it’s not available, the answer would be:

The average value of load current ((I_{avg})) in a type A step down chopper can be calculated using the following formula:

[

I_{avg} = frac{V_{o}}{R}

]

Where:

– (V_{o} = D cdot V_{s} – E) (average output voltage)

– (D) is the duty cycle (in decimal form)

– (V_{s}) is the input voltage (220 V)

– (E) is the back EMF (24 V)

– (R) is the load resistance (1 Ω)

Given:

– (D = 0.30)

– (V_{s} = 220 , V)

– (E = 24 , V)

– (R = 1 , Omega)

First, calculate (V_{o}):

[

V_{o} = D cdot V_{s} – E

]

[

V_{o} = 0.30 cdot 220 , V – 24 , V

]

[

V_{o} = 66 , V – 24 , V = 42 , V

]

Now calculate the average load current:

[

I_{avg} = frac{V_{o}}{R} = frac{42 , V}{1 ,

To find the value of the firing angle ( alpha ) for which the thyristor current ( I ) is maximum, we start with the given expression:

[

I = frac{alpha^2 V_s – alpha E}{R}

]

To find the maximum current with respect to ( alpha ), we can take the derivative of ( I ) with respect to ( alpha ) and set it to zero:

[

frac{dI}{dalpha} = frac{2alpha V_s – E}{R} = 0

]

Setting the derivative equal to zero gives:

[

2alpha V_s – E = 0

]

Solving for ( alpha ) gives:

[

2alpha V_s = E implies alpha = frac{E}{2V_s}

]

Thus, the value of the firing angle ( alpha ) for which the thyristor current is maximum is:

[

alpha = frac{E}{2V_s}

]