Question

decrease increase not change
figure 2 - 28
if the turns ratio of the transformer in figure 2 - 28 is decreased, the forward current through the diode will

Explanation:

Step1: Recall transformer voltage - current relationship

The turns ratio formula is \( \frac{V_1}{V_2}=\frac{N_1}{N_2}\) (where \(V\) is voltage and \(N\) is the number of turns). Also, for an ideal transformer (neglecting losses), \(V_1I_1 = V_2I_2\). The output voltage \(V_{out}\) (across \(R_L\)) is related to the secondary - side voltage of the transformer. The diode \(D_1\) conducts when the secondary - side voltage is positive (for a half - wave rectifier, which this appears to be). The forward current through the diode \(I_D=\frac{V_{out}}{R_L}\) (assuming ideal diode, \(V_{D - on}=0\)).

Step2: Analyze the effect of decreasing turns ratio

Given the turns ratio \(a=\frac{N_1}{N_2}\). If \(a\) is decreased (i.e., \(N_2\) increases while \(N_1\) is constant or \(N_1\) decreases and \(N_2\) is constant in a way that \(\frac{N_1}{N_2}\) gets smaller). From \(V_2=\frac{N_2}{N_1}V_1\), when \(a = \frac{N_1}{N_2}\) decreases, \(V_2\) (the secondary - side voltage) increases. Since \(I_D=\frac{V_{out}}{R_L}\) and \(V_{out}\) (which is equal to the secondary - side voltage for a half - wave rectifier with ideal diode) is directly proportional to \(V_2\), as \(V_2\) increases, the forward current through the diode \(I_D\) (using \(I_D=\frac{V_{out}}{R_L}\), with \(R_L\) constant) will increase.

Answer:

increase