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Nodal analysis voltage drop on 8Ω

Short answer

Nodal analysis with nodes A, B, C and solving the system yields VA = 2.304 V, VB = 2.087 V, VC = −4.043 V. Voltage across the 8 Ω resistor is |(VA − 8 − VC)| = 1.652 V, rounded to 1.65 V.

Solve the given network with nodal analysis to find voltage drop on 8Ω resistor. [Figure: 1 A current source and 2 Ω on the left; 8 Ω in series with an 8 V source on the top; two 4 Ω resistors; an 8 V source with 4 Ω in the middle branch; 6 Ω and a 2 A current source on the right.]

Reading the figure

Bottom rail = ground. Three nodes:

  • A (left): 1 A source into A, 2 Ω to ground, 4 Ω to B, and the branch 8 Ω in series with an 8 V battery (+ on the 8 Ω side) to C.
  • B (middle): 4 Ω to A, 4 Ω to C, and the branch 8 V battery (+ at B) in series with 4 Ω to ground.
  • C (right): 6 Ω to ground, 2 A source leaving C towards ground.

Nodal equations (currents leaving each node)

Node A: V_A/2 + (V_A − 8 − V_C)/8 + (V_A − V_B)/4 = 1 Node B: (V_B − V_A)/4 + (V_B − V_C)/4 + (V_B − 8)/4 = 0 Node C: (V_C − V_A + 8)/8 + (V_C − V_B)/4 + V_C/6 + 2 = 0

Solving: V_A = 2.304 V, V_B = 2.087 V, V_C = −4.043 V

Voltage across the 8 Ω resistor

I₈ = (V_A − 8 − V_C)/8 = (2.304 − 8 + 4.043)/8 = −0.2065 A

V₈ = 8 × 0.2065 = 1.652 V (38/23 V)

The current flows from C towards A through the 8 Ω.

Answer: Voltage drop across 8 Ω ≈ 1.65 V

Assumption: battery polarities taken from the plate lengths in the figure.

✓ Answer checked by Claude Opus 5.5 (high effort)

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