Determine $I_L$ by using Norton’s theorem.

Question

The superposition theorem is essentially based on the concept of:

Answer 1

Step 1: Short terminals A-B to get the Norton current: the source sees $j1\,\Omega$ in series with the two $1\,\Omega$ resistors in parallel $(0.5\,\Omega)$, giving a total of $0.5+j1\,\Omega$ and a source current $\dfrac{20\angle 0^\circ}{0.5+j1} = 17.88\angle -63.43^\circ$ A, which splits equally between the two 1 Ω branches to give $I_N = 8.94\angle -63.43^\circ$ A.
Step 2: With the source de-activated, the Norton impedance seen from A-B is $Z_N = 1 + (1 \parallel j1) = 1.5+j0.5\,\Omega$.
Step 3: With the capacitive load $Z_L=-j1\,\Omega$ reconnected, current division gives $I_L = I_N \times \dfrac{Z_N}{Z_N+Z_L} = 8.94\angle -63.43^\circ \times \dfrac{1.5+j0.5}{1.5-j0.5}$.
\[ \boxed{I_L = 8.94\angle -26.56^\circ \text{ A}} \]

Answer 2

A third way to reach the same load current is to skip forming any Thevenin or Norton equivalent altogether and instead write mesh (loop) equations directly on the original circuit with the capacitive load left in place, then solve for the loop current that flows through it.

Assign a loop current $I_1$ through the source, the series inductive branch and one of the $1\,\Omega$ resistors, and a loop current $I_2$ through the other $1\,\Omega$ resistor and the capacitive load, sharing the middle branch. Writing Kirchhoff's voltage law for each loop and substituting the same component values used earlier reduces to the same two simultaneous equations that the Norton/Thevenin reduction is built to simplify, and solving them for the current through the $-j1\,\Omega$ branch reproduces: \[ I_L = 8.94\angle -26.56^\circ \text{ A} \]

$5.5\angle 22.45^\circ$ A: Solving the loop equations directly does not produce this magnitude or this sign of angle.
$6.5\angle -22.45^\circ$ A: The sign of the angle is right in direction, but the magnitude from the mesh solution is higher than 6.5 A.
$8.94\angle -26.56^\circ$ A: This is exactly what solving the mesh equations for the load branch current gives.
$7.5\angle 26.56^\circ$ A: The angle here has the wrong sign (leading, not lagging) compared with the mesh solution.

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

Answer 3

A third way to reach the same load current is to skip forming any Thevenin or Norton equivalent altogether and instead write mesh (loop) equations directly on the original circuit with the capacitive load left in place, then solve for the loop current that flows through it.

Assign a loop current $I_1$ through the source, the series inductive branch and one of the $1\,\Omega$ resistors, and a loop current $I_2$ through the other $1\,\Omega$ resistor and the capacitive load, sharing the middle branch. Writing Kirchhoff's voltage law for each loop and substituting the same component values used earlier reduces to the same two simultaneous equations that the Norton/Thevenin reduction is built to simplify, and solving them for the current through the $-j1\,\Omega$ branch reproduces: \[ I_L = 8.94\angle -26.56^\circ \text{ A} \]

$5.5\angle 22.45^\circ$ A: Solving the loop equations directly does not produce this magnitude or this sign of angle.
$6.5\angle -22.45^\circ$ A: The sign of the angle is right in direction, but the magnitude from the mesh solution is higher than 6.5 A.
$8.94\angle -26.56^\circ$ A: This is exactly what solving the mesh equations for the load branch current gives.
$7.5\angle 26.56^\circ$ A: The angle here has the wrong sign (leading, not lagging) compared with the mesh solution.

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

Determine $I_L$ by using Norton’s theorem.

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The Correct Option is C

Approach Solution - 1

Approach Solution -2

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