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The average value for the saw-tooth volt...

The average value for the saw-tooth voltage of peak value of `V_(0)` over half the cycle as shown in figure is

A

`(V_(0))/(sqrt(3))`

B

`(V_(0))/(sqrt(2))`

C

`(2V_(0))/(3)`

D

`(V_(0))/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D

`V_("mean") = (int v dt)/(int dt), V = (2V_(0))/(T) t - V_(0)`
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