Question 2.10: Derive the Thevenin Equivalent for the circuit given below....
Derive the Thevenin Equivalent for the circuit given below.
The requirement “Thevenin Equivalent” implies the determination of V_{Th} and R_{Th}. Combination of these two key components, as depicted in diagram below, consti-tutes the Thevenin Equivalent for the given circuit.
First, let’s focus on the calculation of V_{Th}, the voltage shown in the diagram below by applying Ohm’s law.
I=\frac{50 V-V_{Th}}{2 Ω}{Ohm’s Law};or,V_{Th}=50 V-(2 Ω).I
And, since there is an open circuit between point A and Ground,
I=\frac{50 V-20 V}{6 Ω}{Ohm’s Law}=5 A
Then,V_{Th}=50 V-(2 Ω)⋅(5 A)=40 V
Next, we focus on R_{Th}. To determine the value of R_{Th}, we view the circuit from the vantage point of A and Ground, inward. Although some might find this observa-tion less than obvious, as you look into the A – B (A to Ground) plane, you find that 2 Ω and the 4 Ω resistors are in parallel. You can visualize this by imagining that you crease the page along the line formed by points A, B, and Ground. Also, in accordance with the principles of Thevenin’s Theorem, when computing R_{Th}, you neutralize the 50 V and 20 V voltage sources. Hence, the parallel combination of the 2 Ω and the 4 Ω resistors, as shown below, would result in R_{Th} of 1.33 Ω.
R_{Th}=\frac{(2 Ω)(4 Ω)}{2 Ω+4 Ω}=1.33 ΩThe Thevenin Equivalent would then appear as follows:
