In the figure given, the electric current flowing through the 5 k$$\Omega$$ resistor is $$x$$ mA.

The value of $$x$$ to the nearest integer is ________.

We need to find the value of $$x$$ to the nearest integer, which represents the electric current flowing through the $$5\text{ k}\Omega$$ resistor in the given circuit.

Based on the circuit configuration, the system consists of a $$21\text{ V}$$ battery with an internal resistance of $$1\text{ k}\Omega$$, connected in series with a $$5\text{ k}\Omega$$ resistor and a parallel combination of three identical $$3\text{ k}\Omega$$ resistors.

1. Find the Equivalent Resistance of the Parallel Section ($$R_p$$)

The three $$3\text{ k}\Omega$$ resistors are connected in parallel. Their equivalent resistance is calculated as follows:

$$\frac{1}{R_p} = \frac{1}{3\text{ k}\Omega} + \frac{1}{3\text{ k}\Omega} + \frac{1}{3\text{ k}\Omega} = \frac{3}{3\text{ k}\Omega} = 1\text{ k}\Omega^{-1}$$

$$R_p = 1\text{ k}\Omega$$

2. Calculate the Total Equivalent Resistance of the Circuit ($$R_{total}$$)

The internal resistance of the battery ($$1\text{ k}\Omega$$), the main resistor ($$5\text{ k}\Omega$$), and the parallel network branch ($$R_p = 1\text{ k}\Omega$$) are all connected in series. Therefore, the total resistance of the loop is:

$$R_{total} = R_{internal} + R_{5\text{k}} + R_p$$

$$R_{total} = 1\text{ k}\Omega + 5\text{ k}\Omega + 1\text{ k}\Omega = 7\text{ k}\Omega = 7 \times 10^3\ \Omega$$

3. Determine the Main Circuit Current ($$I$$)

Using Ohm's law, the total current leaving the battery and passing directly through the series components (including the $$5\text{ k}\Omega$$ resistor) is:

$$I = \frac{V}{R_{total}} = \frac{21\text{ V}}{7 \times 10^3\ \Omega} = 3 \times 10^{-3}\text{ A} = 3\text{ mA}$$

Since the $$5\text{ k}\Omega$$ resistor is in the main series path, the entire current of $$3\text{ mA}$$ flows through it. Comparing this with the given expression $$x\text{ mA}$$, we get $$x = 3$$.

Therefore, the value of $$x$$ to the nearest integer is 3.