Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man
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Answer:
Option C is correct.
Vs = 9 V, Rth = 30 ohms
Explanation:
In the circuit diagram, there are two sources, so, using superposition, we'll find voltage across A and B, Vab, with respect to each of the sources.
Taking the voltage source as primary source.
We will open circuit the current source just like it is presented in the first drawing of the 2nd image I have added to this solution.
By open-circuiting the current source, the three resistors R₁, R₂ and R₃ are in series.
This means that the Vab is the voltage across the R₃ resistor.
Using voltage divider rule,
Vab = (R₃/R₁ + R₂ + R₃) Vs₁
Vab = [40/(60 + 60 + 40)] (12)
Vab = 3 V
Taking the current source as the primary source.
We will now short circuit the voltage source as shown in the 2nd drawing of the 2nd image I have attached to this solution.
As shown in the 2nd drawing of the 2nd file I attached to this solution, the R₂ is in parallel with R₁ and R₃, that is R₂//(R₁ + R₃)
Using the current divider rule
Current in the (R₁ + R₃) branch = [R₂/(R₂ + (R₁ + R₃))] Is₂ = (60/(60 + 60 + 40)) (0.4) = 0.15 A
Vab = voltage across the R₃ resistor = IR₃ = 0.15 × 40 = 6 V
Total voltage across A and B, Vab = Vab due to Vs₁ + Vab due to Is₂ = 3 + 6 = 9 V
And for the Rth, we open-circuit the current source and short-circuit the voltage source simultaneously and look at the resistance of the setup from the AB terminal, as shown in the first drawing of the 3rd file attached to this solution.
It is evident that R₃ is in a parallel combination with (R₁ + R₂)
Rth = (R₁ + R₂)//R₃ = (60 + 60)//40 = 120//40 = (120× 40)/(120 + 40) = 30 ohms
The image of the thevenin equivalent of the circuit as seen from terminals AB is presented in the 2nd drawing on the 3rd image attached.
