In the circuit diagram shown in figure given below, the current flowing through resistance 3 $$\Omega$$ is $$\frac{x}{3}$$ A. The value of $$x$$ is _______.

Let the 6 V cell and the 3 Ω resistor in the left-hand branch constitute loop 1. Let the 3 V cell and the 3 Ω resistor in the right-hand branch constitute loop 2. The upper ends of the two branches are joined by a 3 Ω resistor (the one in which the current is asked).

Assign clockwise loop currents:   • $$i_1$$ flows in loop 1 (through the left 3 Ω resistor and then through the 3 Ω junction resistor from left to right).
  • $$i_2$$ flows in loop 2 (through the right 3 Ω resistor and then through the same junction resistor from right to left).
Hence, through the 3 Ω junction resistor the actual current is $$i_1 - i_2$$ from left to right.

Case 1: KVL in loop 1
Starting from the negative terminal of the 6 V cell and moving clockwise, the potential rises by 6 V and then drops across the two resistors:
$$6 - 3\,i_1 - 3\,(i_1 - i_2) = 0$$

Simplifying, $$6 - 3i_1 - 3i_1 + 3i_2 = 0$$ $$6 - 6i_1 + 3i_2 = 0$$ Divide by 3: $$2 - 2i_1 + i_2 = 0 \qquad -(1)$$

Case 2: KVL in loop 2
Going clockwise round loop 2 gives a rise of 3 V (the 3 V cell) followed by drops across its two resistors:
$$3 - 3\,i_2 - 3\,(i_2 - i_1) = 0$$

Simplifying, $$3 - 3i_2 - 3i_2 + 3i_1 = 0$$ $$3 + 3i_1 - 6i_2 = 0$$ Divide by 3: $$1 + i_1 - 2i_2 = 0 \qquad -(2)$$

Solving equations (1) and (2)
From $$(1):\; i_2 = 2i_1 - 2$$

Substitute this $$i_2$$ in $$(2):$$ $$1 + i_1 - 2(2i_1 - 2) = 0$$ $$1 + i_1 - 4i_1 + 4 = 0$$ $$5 - 3i_1 = 0$$ $$i_1 = \frac{5}{3}\,\text{A}$$

Now, $$i_2 = 2\left(\frac{5}{3}\right) - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3}\,\text{A}$$

Current through the 3 Ω junction resistor
$$i = i_1 - i_2 = \frac{5}{3} - \frac{4}{3} = \frac{1}{3}\,\text{A}$$

The current is given in the question as $$\dfrac{x}{3}\,\text{A}$$. Comparing, $$x = 1$$.

Therefore, the required value of $$x$$ is 1.