Find v_L for the full-wave rectifier circuit of Fig. 2-43(*a*), treating the transformer and diodes as ideal. Assume R_S = 0.

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The two voltages labeled v_2 in Fig. 2-43(*a*) are identical in magnitude and phase. The ideal transformer and the voltage source v_S can therefore be replaced with two identical voltage sources, as in Fig. 2-43(*b*), without altering the electrical performance of the balance of the network. When v_S/n is positive, D_1 is forward-biased and conducts but D_2 is reverse-biased and blocks. Conversely, when v_S/n is negative, D_2 conducts and D_1 blocks. In short,

i_{D1} = \begin{cases} \frac{v_S/n}{R_L} \frac{v_s}{n} \geq 0 \\ 0\quad\quad\frac{v_s}{n} \lt 0 \end{cases} \quad \text{and} \quad i_{D2} = \begin{cases} 0\quad\quad\frac{v_s}{n} \gt 0 \\ -\frac{v_S/n}{R_L} \frac{v_s}{n} \leq 0 \end{cases}

By KCL, i_L = i_{D1} + i_{D2} = \frac{|v_s/n|}{R_L}

and so v_L = R_L i_L = |v_s/n|.

Question: 2.2