Question

Determine $I_L$ by using Thevenin’s theorem. 

• A. 3.5 A
• B. 2.5 A
• C. 4 A
• D. 5 A

Hint:

While applying Thevenin’s theorem, always remove the load first, calculate open-circuit voltage for $V_{th}$, and deactivate sources properly to find $R_{th}$.

Answer 1

The Correct Option is D

Step 1: Remove the 1 Ω load and reduce the source network to its Thevenin equivalent: analysis of the two branches gives \( V_{th} = 20 \) V and \( R_{th} = 3\,\Omega \) at terminals A-B.
Step 2: Reconnect the 1 Ω load in series with this equivalent, so the total loop resistance is \( 3 + 1 = 4\,\Omega \).
Step 3: Apply Ohm's law to the single-loop equivalent circuit: \( I_L = \dfrac{V_{th}}{R_{th}+R_L} = \dfrac{20}{4} \).
\[ \boxed{I_L = 5 \text{ A}} \]

Answer 2

Rather than reducing the network to a Thevenin or Norton equivalent, the load current can be obtained directly by writing a node-voltage equation at the junction node with the 1 Ω load left in place. Since the resistive network reduces, once the load is included, to the same effective 4 Ω seen by the combined source contributions established above (3 Ω of internal resistance in series with the 1 Ω load), solving the single node equation for the current drawn from the equivalent 20 V drive gives the same relationship \( I_L = V_{th}/(R_{th}+R_L) \) directly, without separately packaging the network as a labelled Thevenin or Norton source.

1. 3.5 A: Solving the node equation with the actual 20 V, 3 Ω and 1 Ω values present in the circuit does not produce this figure.
2. 2.5 A: This would only result if the effective source resistance were twice as large as the 3 Ω obtained from the network reduction, which is not the case here.
3. 4 A: This would follow only if the load resistor were 2 Ω, not the 1 Ω actually specified.
4. 5 A: Solving \( \dfrac{20}{3+1} \) directly from the node equation gives exactly this value.

Therefore, the correct answer is 5 A.