Question

Determine $I_L$ by using Norton’s theorem. 

• A. $5.5 \angle 22.45^\circ$ A
• B. $6.5 \angle -22.45^\circ$ A
• C. $8.94 \angle -26.56^\circ$ A
• D. $7.5 \angle 26.56^\circ$ A

Hint:

For AC circuits, Norton’s theorem is applied using complex impedances. Always calculate Norton current using short-circuit conditions and include phase angles carefully.

Answer 1

The Correct Option is C

Step 1: Identify the load terminals A–B.
The load is the capacitive reactance $-j\,\Omega$ connected between terminals A and B. To apply Norton’s theorem, this load is first removed.
Step 2: Calculate Norton current $I_N$.
Norton current is the short-circuit current between terminals A and B. Hence, terminals A and B are shorted.
With A–B shorted, the $1\,\Omega$ series resistor is directly connected to ground, making the circuit equivalent to a source feeding a parallel combination.
Impedances seen from the source are:
Inductive reactance: $j1\,\Omega$
Parallel branch: $1\,\Omega \parallel 1\,\Omega = 0.5\,\Omega$
Total impedance:
\[ Z_{\text{total}} = 0.5 + j1 \]
Source current is:
\[ I = \frac{20\angle 0^\circ}{0.5 + j1} \]
\[ |Z| = \sqrt{0.5^2 + 1^2} = 1.118,\quad \angle Z = 63.43^\circ \]
\[ I = \frac{20}{1.118} \angle (-63.43^\circ) = 17.88 \angle -63.43^\circ \text{ A} \]
Since the current divides equally between the two parallel $1\,\Omega$ branches,
\[ I_N = \frac{17.88}{2} = 8.94 \angle -63.43^\circ \text{ A} \]
Step 3: Calculate Norton impedance $Z_N$.
Deactivate the source by replacing the voltage source with a short circuit.
Seen from terminals A–B:
\[ Z_N = 1 + (1 \parallel j1) \]
\[ 1 \parallel j1 = \frac{1 \cdot j1}{1 + j1} = 0.5 + j0.5 \]
\[ Z_N = 1.5 + j0.5 \]
Step 4: Reconnect the load and find $I_L$.
Load impedance:
\[ Z_L = -j1 \]
Using current division:
\[ I_L = I_N \times \frac{Z_N}{Z_N + Z_L} \]
\[ Z_N + Z_L = 1.5 + j0.5 - j1 = 1.5 - j0.5 \]
Magnitude and angle:
\[ \left|\frac{Z_N}{Z_N + Z_L}\right| = 1,\quad \angle = 36.87^\circ \]
Hence,
\[ I_L = 8.94 \angle (-63.43^\circ + 36.87^\circ) \]
\[ I_L = 8.94 \angle -26.56^\circ \text{ A} \]
Step 5: Conclusion.
The load current is
\[ \boxed{8.94 \angle -26.56^\circ \text{ A}} \]