To determine the power loss in the diode, we need to calculate the product of the on-state current and the on-state voltage drop across the diode. The given figure provides details about the diode's on-state characteristics.

From the graph, we observe the following:

• The on-state voltage at 50 A is approximately \(1.8 \, \text{V}\).
• The relationship between current and voltage is linear, so we can use this ratio to find the voltage drop for 15 A.

Using the graph, the equation of the line can be approximated from the two known points: \((0, 1.0)\) V and \((50, 1.8)\) V.

The slope (m) of the line is calculated as:

\(m = \frac{(1.8 - 1.0) \, \text{V}}{(50 - 0) \, \text{A}} = \frac{0.8}{50} = 0.016 \, \text{V/A}\)

Now, we can find the voltage drop (\(V_{\text{ON}}\)) at 15 A:

\(V_{\text{ON}} = 1.0 + 0.016 \times 15 = 1.0 + 0.24 = 1.24 \, \text{V}\)

The power loss (\(P_{\text{loss}}\)) in the diode is given by:

\(P_{\text{loss}} = I_{\text{ON}} \times V_{\text{ON}}\)

\(P_{\text{loss}} = 15 \, \text{A} \times 1.24 \, \text{V} = 18.6 \, \text{W}\)

Therefore, the power loss in the diode is 18.6 W.