To find the current in the given circuit, we need to analyze the diode's operation and apply the given $i-v$ characteristics. The circuit consists of a 10 V battery, a 1 kΩ resistor, and a diode.

The $i-v$ characteristics of the diode are:

• If \( v \ge 0.7 \, \text{V} \), then \( i = \frac{v-0.7}{500} \, \text{A} \)
• If \( v < 0.7 \, \text{V} \), then \( i = 0 \, \text{A} \)

First, let's use Kirchhoff's Voltage Law (KVL) around the loop:

\(10 - i \cdot 1000 - v = 0\)

Rearrange to solve for \( v \):

\(v = 10 - 1000i\)

For the diode to conduct, \( v \) must be at least 0.7 V. Therefore, let:

\(10 - 1000i \ge 0.7\)

Simplifying, we get:

\(1000i \le 9.3\)

\(i \le 0.0093 \, \text{A} \, \text{or} \, 9.3 \, \text{mA}\)

Using the diode's characteristic equation \( i = \frac{v-0.7}{500} \), substitute \( v = 10 - 1000i \)

\(i = \frac{10 - 1000i - 0.7}{500}\)

Multiply throughout by 500 to clear the fraction:

\(500i = 10 - 0.7 - 1000i\)

\(500i + 1000i = 9.3\)

\(1500i = 9.3\)

Solve for \( i \):

\(i = \frac{9.3}{1500} = 0.0062 \, \text{A} \, \text{or} \, 6.2 \, \text{mA}\)

Therefore, the current in the circuit is 6.2 mA.