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
\[ \boxed{5 \text{ A}} \]

Answer 2

Instead of finding the Thevenin voltage by direct voltage division and then simply adding source contributions, the same open-circuit voltage at terminals A-B can be built up by superposition, treating one source at a time, and the load current can then be found through the Norton (current-divider) route rather than a straight Ohm's law step.

With the 10 V source short-circuited and only the 20 V source active, the two 2 Ω resistors on the left form a voltage divider, giving \( 20 \times \frac{2}{2+2} = 10 \) V at the junction node; since no current flows through the open right-hand branch, this 10 V appears unchanged at terminal A. With the 20 V source short-circuited and only the 10 V source active, again no current can flow through the open branch, so the full 10 V of that source appears at terminal A. Adding the two contributions gives \( V_{th} = 10 + 10 = 20 \) V, and deactivating both sources for the resistive reduction gives \( R_{th} = 2 + (2 \parallel 2) = 3\,\Omega \), matching the standard reduction.

Converting to a Norton equivalent, \( I_N = \frac{V_{th}}{R_{th}} = \frac{20}{3} = 6.67 \) A in parallel with \( R_{th} = 3\,\Omega\). Feeding the 1 Ω load, the current divider gives: \[ I_L = I_N \times \frac{R_{th}}{R_{th}+R_L} = 6.67 \times \frac{3}{4} = 5 \text{ A} \]

1. 3.5 A: This would require either a smaller Thevenin voltage or a larger total resistance than the circuit actually has; it does not match the 20 V, 3 Ω, 1 Ω combination found here.
2. 2.5 A: This value would arise only if the total resistance were doubled to 8 Ω, which is not the case here.
3. 4 A: This would follow if the load resistance were 2 Ω instead of 1 Ω, which does not match the given circuit.
4. 5 A: This is exactly what the superposition-plus-Norton route gives, \( \frac{20}{3+1} = 5 \) A.

Therefore, the correct answer is 5 A.