Using Superposition theorem, find voltage across 4 Ω resistor. [Circuit: a common bottom wire (reference). Left vertical branch: a 10 A current source between the bottom wire and the top-left wire, arrow pointing upward (current delivered into the to…

Take the bottom wire as reference (0 V). Let V_X be the voltage of node X = voltage across the 4 Ω resistor.

Step 1: 10 A source acting alone (25 V and 15 V replaced by short circuits). Node Y is shorted to the bottom wire by the shorted 25 V source, so the 1 Ω resistor appears directly across the 4 Ω resistor (and the 3 Ω branch is also shorted out by the 25 V short).

\( R_{eq} = 4\parallel1 = \dfrac{4\times1}{4+1} = 0.8\ \Omega \) ⇒ \( V_1 = 10\times0.8 = 8 \) V (top of 4 Ω positive).

Step 2: 25 V source acting alone (10 A open-circuited, 15 V shorted). Node Y is at +25 V. The path Y → 1 Ω → X → 4 Ω → ground is a voltage divider (the 3 Ω branch is directly across the 25 V source and does not affect X):

\( V_2 = 25\times\dfrac{4}{4+1} = 20 \) V (top positive).

Step 3: 15 V source acting alone (10 A open, 25 V shorted). The 25 V short holds node Y at 0 V, so the 15 V source only drives current through the 3 Ω resistor and the short; no current reaches the 1 Ω / 4 Ω part.

\( V_3 = 0 \) V.

Step 4: Superposition

\( V_{4\Omega} = V_1+V_2+V_3 = 8+20+0 = \) 28 V (top end of 4 Ω positive). Current in 4 Ω = 28/4 = 7 A downward.

Check by nodal analysis: V_Y = 25 V. KCL at X: \( 10 = \dfrac{V_X}{4} + \dfrac{V_X-25}{1} \Rightarrow 35 = 1.25V_X \Rightarrow V_X = 28 \) V. ✓ (The result does not depend on the polarity of the 15 V source.)