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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)` |
Answer» Correct Answer - D `V_("mean") = (int v dt)/(int dt), V = (2V_(0))/(T) t - V_(0)` |
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