The figure shows a straight-line approximation for the forward characteristics of a power diode. A continuous on-state current of 15 A is flowing through the diode. What is the power loss in the diode?

The correct answer is 18.6 W.

Step 1: Extract diode characteristics from the given graph
The provided image shows the piecewise linear (straight-line) approximation of the forward characteristics (On-state current vs. On-state voltage) of the power diode.
From the graph, we can identify two key coordinate points $(V_D,I_D)$:
1. The line intercepts the horizontal axis (voltage) at $V_D=1.0\text{V}$ when the current is $I_D=0\text{A}$. This is the threshold (or cut-in) voltage of the diode:
$V_{T0}=1.0\text{V}$
2. A second point is marked by dashed lines at $V_D=1.8\text{V}$ corresponding to a current of $I_D=50\text{A}$.

Step 2: Calculate the dynamic on-state resistance of the diode
The slope of the characteristic line determines the dynamic resistance ($r_D$):
$r_D=\frac{\Delta V_D}{\Delta I_D}=\frac{1.8\text{V}-1.0\text{V}}{50\text{A}-0\text{A}}$
Evaluating this expression:
$r_D=\frac{0.8\text{V}}{50\text{A}}=0.016\text{Ω}$

Step 3: Determine the forward voltage drop at a continuous current of 15 A
The forward voltage equation representing the straight-line characteristic is:
$V_D=V_{T0}+(I_D\times r_D)$
Substitute the given continuous current $I_D=15\text{A}$:
$V_D=1.0\text{V}+(15\text{A}\times 0.016\text{Ω})$
$V_D=1.0\text{V}+0.24\text{V}=1.24\text{V}$

Step 4: Calculate the power loss in the diode
Since a continuous current is flowing through the diode, the average power loss ($P_{\text{loss}}$) is given by the product of the diode voltage drop and the current:
$P_{\text{loss}}=V_D\times I_D$
$P_{\text{loss}}=1.24\text{V}\times 15\text{A}=18.6\text{W}$