## Answers

Center top full wave Rectifier a) it is given that v_i = 120 v () at 60 K Rightarrow N_1/E_2 = 6/1 Rightarrow R = 2 k Ohm Rightarrow D_1, D_2 where 1N4002 a) sketch waveform at the load v_i/v_s = N_1/N_2 = 6/1 V-S = v_i/6 = 120/6 = 20 v V_S/2 = 10 v () V_ = 10 squareroot 12 v = 14.14 v waveform D_1 turns on the +ve cycle is applied (drop = 0.7 v) when D_1, ON, D_2 is OFF (because -ve is applied) voltage \r\n (b) Average voltage at output V_max = V_P - 0.7 v (dt drop) = (10 squareroot 2 - 0.7) v Average voltage = 1/T integral_0^T V_max sin w t d w t = 1/pi integral_0^pi V_max sin w t d w t = V_max/pi integral_0^T sin w t dt = V_amx/pi - [Cos w t]_0^pi = V_max/pi - [Cos pi - Cos 0] Average voltage - 2 v_max/pi Average voltage = 2 [10 squareroot 2 - 0.7/pi] = 8.55v Answer: Average voltage = 8.55 v

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