Skip to main content

Nodal analysis, currents in 5Ω and 3Ω

Short answer

Nodal equations at L, R, M yield V_L ≈ 15.99 V, V_R ≈ 2.34 V. Then I₅Ω = (15.99−2.34)/5 = 2.73 A; I₃Ω = (15.99−5)/3 = 3.66 A.

Using nodal analysis find the current through the 5 ohm and 3 ohm resistors of the following circuit. [Figure: 6 A current source on top; 5 Ω and 4 Ω and 2 Ω in the upper part; 3 Ω with a 5 V source on the left; 4 A current source in the middle; 1 Ω with a 2 V source on the right.]

Reading the figure

Bottom rail = ground. Three nodes:

  • L (left): 3 Ω in series with the 5 V battery (+ on top) to ground; 5 Ω to R; 2 Ω to M; the 6 A source (arrow left, from R to L) delivers current into L.
  • R (right): 1 Ω in series with the 2 V battery (+ on top) to ground; 4 Ω to M; the 6 A source takes current out of R.
  • M (middle): 2 Ω to L, 4 Ω to R, and the 4 A source (arrow up) injecting current into M from ground.

Nodal equations (currents leaving each node)

Node L: (V_L − 5)/3 + (V_L − V_R)/5 + (V_L − V_M)/2 = 6 Node R: (V_R − 2)/1 + (V_R − V_L)/5 + (V_R − V_M)/4 = −6 Node M: (V_M − V_L)/2 + (V_M − V_R)/4 = 4

Multiply by 30, 20 and 4:

  • 31 V_L − 6 V_R − 15 V_M = 230
  • −4 V_L + 29 V_R − 5 V_M = −80
  • −2 V_L − V_R + 3 V_M = 16

Solving: V_L = 15.99 V, V_R = 2.34 V, V_M = 16.77 V

Required currents

  • 5 Ω: I = (V_L − V_R)/5 = (15.99 − 2.34)/5 = 2.73 A (from L to R)
  • 3 Ω: I = (V_L − 5)/3 = (15.99 − 5)/3 = 3.66 A (from L down into the 5 V battery)

Other branch currents: 2 Ω (L to M) = −0.39 A, 4 Ω (M to R) = 3.61 A, 1 Ω = 0.34 A.

Answer: I₅Ω ≈ 2.73 A; I₃Ω ≈ 3.66 A

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 2 questions →