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
\[ \boxed{8.94 \angle -26.56^\circ \text{ A}} \]

Answer 2

Since Thevenin and Norton equivalents describe the same source network, the load current can equally be found by first converting to the Thevenin form (a series voltage source and impedance) instead of working with the Norton current source and current division.

Using the same Norton current and Norton impedance found by shorting and then de-activating the source, \(I_N = 8.94\angle -63.43^\circ\) A and \(Z_N = 1.5+j0.5\,\Omega\), the equivalent Thevenin voltage is \(V_{th} = I_N Z_N\). Since \(|Z_N| = \sqrt{1.5^2+0.5^2} = 1.58\,\Omega\) at an angle of \(\tan^{-1}(0.5/1.5) = 18.43^\circ\): \[ V_{th} = 8.94 \angle -63.43^\circ \times 1.58\angle 18.43^\circ = 14.14 \angle -45^\circ \text{ V} \]

Reconnecting the capacitive load \(Z_L = -j1\,\Omega\) in series with \(Z_{th} = 1.5+j0.5\,\Omega\) and applying Ohm's law directly: \[ Z_{th}+Z_L = 1.5 - j0.5, \qquad |Z_{th}+Z_L| = 1.58\,\Omega \text{ at } -18.43^\circ \] \[ I_L = \frac{14.14\angle -45^\circ}{1.58\angle -18.43^\circ} = 8.94 \angle (-45^\circ+18.43^\circ) = 8.94\angle -26.57^\circ \text{ A} \]

1. \(5.5\angle 22.45^\circ\) A: Neither the magnitude nor the sign of the angle obtained from the Thevenin route matches this option.
2. \(6.5\angle -22.45^\circ\) A: The angle is close in sign but the magnitude from the calculation above is noticeably larger than 6.5 A.
3. \(8.94\angle -26.56^\circ\) A: This matches the magnitude and angle obtained from the Thevenin-based calculation.
4. \(7.5\angle 26.56^\circ\) A: The magnitude of this angle matches but its sign is opposite (leading rather than lagging), which does not agree with the calculation.

Therefore, the correct answer is \(8.94\angle -26.56^\circ\) A.