2 + E1 - b R₁ = 2.2 k E2 + d lx R₂ = ↓ 3.3 ΚΩ R3 = 2.0 ΚΩ Vx Figure 3.0: Circuit with multiple voltage sources

Introductory Circuit Analysis (13th Edition)
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(i) For the circuit shown in Figures 3.0, consider the input D.C. source voltages, E1 = 5V and E2 = 2V (note the polarities of each source!). Determine the values of voltage, VX and current, IX using either  the nodal-voltage or the mesh-current analysis technique. Show your analysis below, and record  your results in Table 4.0.  

 

(ii) For the same circuit shown in Figures 3.0 with the input D.C. source voltages, E1 = 5V and E2 =  2V, apply the Superposition Principle technique as follows:- 

  1. Remove input source, E2 and replace it with a short-circuit by connecting a wire between  “c” and “d”. Determine the resultant voltage, VX1 across R3 and current, IX1 through R3.  Show your analysis below, and record the results in Table 4.0
  2. Connect input source, E2 in its original place between “c” and “d”. Be mindful of the  polarity connections. Then remove input source, E1 and replace it with a short-circuit by  connecting a wire between “a” and “b”. Determine the resultant voltage, VX2 across R3 and  current, IX2 through R3. Show your analysis below, and record the results in Table 4.0
2
+
E1
-
b
R₁ = 2.2 k
E2
+
d
lx
R₂ =
↓
3.3 ΚΩ
R3 =
2.0 ΚΩ
Vx
Figure 3.0: Circuit with multiple voltage sources
Transcribed Image Text:2 + E1 - b R₁ = 2.2 k E2 + d lx R₂ = ↓ 3.3 ΚΩ R3 = 2.0 ΚΩ Vx Figure 3.0: Circuit with multiple voltage sources
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