Find current flowing through 1 Ω resistance by Node analysis. [Circuit: a common bottom wire (reference). Left vertical branch: a 4 V battery from the bottom wire up to the left node L (longer plate, +, at the top). From L the left vertical wire cont…

Reference: bottom wire = 0 V. Node voltages: V_L, V_M, V_R.

The 4 V source is directly between L and reference ⇒ V_L = 4 V.

KCL at node R (currents entering = leaving): 1 A (from the top source) + 3 A (from the right source) enter R; current leaves R through 3 Ω towards M:

\( 1 + 3 = \dfrac{V_R - V_M}{3} \Rightarrow V_R - V_M = 12 \) … (1)

KCL at node M (currents entering from 2 Ω and 3 Ω = current leaving through 1 Ω):

\( \dfrac{V_L-V_M}{2} + \dfrac{V_R-V_M}{3} = \dfrac{V_M}{1} \)

Using (1): \( \dfrac{4-V_M}{2} + 4 = V_M \Rightarrow 2 - 0.5V_M + 4 = V_M \Rightarrow 1.5V_M = 6 \Rightarrow V_M = 4 \) V, and V_R = 16 V.

Current through 1 Ω: \( I = \dfrac{V_M}{1} = \) 4 A, flowing downward (from node M to the bottom wire).

Check: I(2 Ω) = (4 − 4)/2 = 0; I(3 Ω) = (16 − 4)/3 = 4 A into M; 0 + 4 = 4 A leaves through 1 Ω ✓. At R: 1 + 3 = 4 A ✓.