A delayed full-wave rectified sinusoidal current has an average value equal to half its maximum value. Find the delay angle θ.

F1 U.B 18.9.20 Pallavi D14

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  1. \(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2} - \frac{1}{2}} \right)\;\;\)
  2. \(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2} + 1} \right)\)
  3. \(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2} - 1} \right)\)
  4. \(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2}} \right)\)

Answer (Detailed Solution Below)

Option 3 : \(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2} - 1} \right)\)
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Detailed Solution

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Concept:

The average output voltage of a full-wave controlled rectifier with R load is given by:

\({V_0} = \frac{{{V_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Where Vm is the maximum value of supply voltage

α is the firing angle or delay angle

Average load current:

\({I_0} = \frac{{{V_m}}}{{\pi R}}\left( {1 + \cos \alpha } \right) = \frac{{{I_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Calculation:

Given that, the average value of the average load current is equal to half of its maximum value.

\(\frac{{{I_m}}}{2} = \frac{{{I_m}}}{\pi }\left( {1 + \cos \theta } \right)\)

\(\theta = {\cos ^{ - 1}}\left( {\frac{\pi }{2} - 1} \right)\)
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