Determine Norton's equivalent circuit which may be used to represent the given network at the terminals AB and hence find current through 5 Ω resistance. [Circuit: a common bottom wire ending at terminal B. Left branch: a 12 V source (circle, + at th…

Step 1: Norton current I_N (remove 5 Ω, short A–B). The 4 Ω is then in parallel with 6 Ω across N–B.

\( R_{total} = 2 + (6\parallel4) = 2 + 2.4 = 4.4\ \Omega \); source current \( = \dfrac{12}{4.4} = 2.727 \) A.

\( V_N = 2.727\times2.4 = 6.545 \) V ⇒ \( I_N = \dfrac{6.545}{4} = \) 1.636 A (= 18/11 A), from A to B through the short.

Step 2: Norton resistance R_N (12 V shorted, looking into A–B): \( R_N = 4 + (2\parallel6) = 4 + 1.5 = \) 5.5 Ω.

Norton equivalent: 1.636 A current source in parallel with 5.5 Ω across A–B.

    +-------+------ A
    |       |   |
 (1.636 A) 5.5  5 ohm
    | up    |   |
    +-------+------ B

Step 3: current through 5 Ω: \[ I_{5\Omega} = I_N\frac{R_N}{R_N + 5} = 1.636\times\frac{5.5}{10.5} = \mathbf{0.857\ A} \;(= 6/7\ \text{A}) \] from A to B.

Check (Thevenin): V_th = 12 × 6/(2 + 6) = 9 V; I = 9/(5.5 + 5) = 0.857 A ✓; V_th/R_N = 9/5.5 = 1.636 A = I_N ✓.