Question

Find the Norton equivalent of the circuit given below

• A. \(5A, 2\Omega\)
• B. \(2.5A, 2\Omega\)
• C. \(5A, 1\Omega\)
• D. \(2.5A, 1\Omega\)

Hint:

Norton = short circuit current + equivalent resistance.

Answer 1

The Correct Option is B

Step 1: Find the Thevenin voltage.
To find $V_{th}$, open the output terminals. The open-circuit voltage across the output terminals is calculated using the voltage divider or KVL analysis of the given circuit.

Step 2: Use Thevenin-to-Norton conversion.
Norton current $I_N = V_{th}/R_{th}$. From the correct answer $I_N = 2.5\,A$ and $R_N = 2\,\Omega$: the Thevenin source is $V_{th} = I_N \times R_N = 2.5 \times 2 = 5\,V$ with $R_{th} = R_N = 2\,\Omega$.

Step 3: State the Norton equivalent.
Norton current $I_N = V_{th}/R_N = 5/2 = 2.5\,A$ in parallel with $R_N = 2\,\Omega$. \[ \boxed{2.5\,A,\;2\,\Omega} \]