The net current flowing in the given circuit is_______ A.
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The circuit contains only one closed loop, so we can apply Kirchhoff’s Voltage Law (KVL) directly.
KVL statement: “The algebraic sum of the emfs in any closed loop is equal to the algebraic sum of the potential drops (IR terms) in that loop.”

Let the two ideal sources be
  • $$E_1 = 12\;\text{V}$$ (positive terminal on the left side of the loop)
  • $$E_2 = 6\;\text{V}$$ (connected so that it opposes $$E_1$$)
and let the two resistors be
  • $$R_1 = 2\;\Omega$$,  • $$R_2 = 4\;\Omega$$, both in series with the batteries.

Assume the current $$I$$ flows clockwise, starting from the positive terminal of $$E_1$$. Moving once around the loop and writing the KVL equation:

Clockwise sum of emfs = $$E_1 - E_2$$ (because $$E_2$$ is encountered from + to −, so it is subtractive).
Clockwise sum of drops = $$I (R_1 + R_2)$$.
Therefore, by KVL
$$E_1 - E_2 = I (R_1 + R_2) \quad -(1)$$

Substituting the given numerical values into $$(1)$$:

$$12\;{\text{V}} - 6\;{\text{V}} = I (2\;\Omega + 4\;\Omega)$$

$$6\;{\text{V}} = I \,(6\;\Omega)$$

$$I = \frac{6\;\text{V}}{6\;\Omega} = 1\;\text{A}$$

Hence, the net current in the circuit is $$1\;\text{A}$$, flowing clockwise (from the 12 V source towards the 6 V source).