Introduction:
In a Junction Field Effect Transistor (JFET), the drain current (ID) is directly proportional to the gate-to-source voltage (VGS) when it is less than the pinch-off voltage (VP). However, when VGS is equal to or greater than VP, the JFET operates in the saturation region where ID remains relatively constant. In this question, we are given that VGS is 65% of VP and we need to find the percentage of IDSS (maximum drain current) in this case.

Explanation:
To solve this problem, we need to understand the relationship between VGS and ID in a JFET.

1. VGS and ID Relationship:
In the linear region of operation (VGS < vp),="" the="" drain="" current="" is="" given="" by="" the="" />

ID = IDSS * (1 - (VGS / VP))^2

Where:
IDSS is the maximum drain current when VGS = 0
VGS is the gate-to-source voltage
VP is the pinch-off voltage

2. Given Information:
We are given that VGS is 65% of VP, which can be expressed as:

VGS = 0.65 * VP

3. Finding ID:
Using the equation for drain current in the linear region, we substitute the given value of VGS:

ID = IDSS * (1 - (0.65 * VP / VP))^2

Simplifying the expression:

ID = IDSS * (1 - 0.65)^2

ID = IDSS * (0.35)^2

ID = IDSS * 0.1225

4. Calculating the Percentage of IDSS:
To find the percentage of IDSS, we divide ID by IDSS and multiply by 100:

Percentage of IDSS = (ID / IDSS) * 100

Percentage of IDSS = (IDSS * 0.1225 / IDSS) * 100

Percentage of IDSS = 12.25%

Conclusion:
Therefore, the percentage of current IDSS for a JFET, when the gate-to-source voltage is 65% of the pinch-off voltage, is 12.25%. This falls within the range of 12-12.5% as stated in the correct answer.