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 load resistance to find Thevenin voltage.
To determine the Thevenin equivalent as seen from terminals A–B, the load resistance of $1\,\Omega$ is removed. The circuit is now open at terminals A–B.
Step 2: Calculate the Thevenin voltage $V_{th$.}
With the load open, no current flows through the right-side $2\,\Omega$ resistor and the $10\,\text{V}$ source.
The left part of the circuit consists of a $20\,\text{V}$ source feeding two $2\,\Omega$ resistors, one in series and one to ground.
Using voltage division, the voltage at the junction node is
\[ V_N = 20 \times \frac{2}{2+2} = 10 \text{ V} \]
The $10\,\text{V}$ source raises the voltage further at terminal A, hence
\[ V_{th} = 10 + 10 = 20 \text{ V} \]
Step 3: Find the Thevenin resistance $R_{th$.}
Deactivate all independent sources:
- Replace the $20\,\text{V}$ source by a short circuit
- Replace the $10\,\text{V}$ source by a short circuit
Now, from terminal A, the $2\,\Omega$ resistor is in series with two $2\,\Omega$ resistors connected in parallel to ground.
\[ 2 \parallel 2 = 1 \,\Omega \]
Therefore,
\[ R_{th} = 2 + 1 = 3 \,\Omega \]
Step 4: Reconnect the load resistance and calculate $I_L$.
The total resistance seen by the Thevenin voltage source is
\[ R_{\text{total}} = R_{th} + R_L = 3 + 1 = 4 \,\Omega \]
Hence, the load current is
\[ I_L = \frac{V_{th}}{R_{\text{total}}} = \frac{20}{4} = 5 \text{ A} \]
Step 5: Conclusion.
The current flowing through the load resistance is
\[ \boxed{5 \text{ A}} \]