We can also solve this question using laplace transformation..
If you know laplace transformation then I will solve by laplace transform...
Solve by MATLAB, and PSPICE at omega = 1 rad/s. Prob#9 Find Vo by nodal analysis....
solve for v0 using mesh analysis. assume omega = 4000
rad/s
1) What is v0?
2) How do you convert 36sinwt A to polar form?
3. Solve for Vo using mesh analysis. Assume ω = 4000 rad/s.
Find Vo by nodal analysis 1Ω 3Ω 310
please solve USING NODAL ANALYSIS ON BOTH (nodes)SIDES OF
VS
Figure 9.60 For Prob. 9.53 9.54 In the circuit of Fig. 9.61, find V, if I = 30/0° A. ML - 22 Vs - 122 2's 403 j2828 3152 Figure 9.61 For Prob. 9.54.
7. Use nodal analysis to solve for the nodal voltages in the following circuit: 2Α (1) WW 2Ω Μν 4Ω V V 4 Ω V. 2Ω W ΑΛΛ 8Ω ΑΛ 31, 3Α Η
7. Find Vo in the network shown in Figure below using nodal analysis (20 points) 12 V ww 1 k2 1 kO 2 k2 Ans: Use superposition and you should get: VO= 9 V
For the circuit shown above, solve for Vo and io.
PLEASE USE NODAL or SUPERNODAL ANALYSIS. PLEASE DO NOT USE
MESH.
24 12 32 12 + 16 - 2 642 + 4i, 40 V + 5512 0.246 1.1A For the circuit shown above, solve for vo and io.
help
3.7 Apply nodal analysis to solve for V, in the circuit of Fig. 3.56. 2A 1022 www 2002 0.2V Figure 3.56 For Prob. 3.7. 3.12 Using nodal analysis, determine V. in the circuit in Fig. 3.61. 10 22 2022 NW + 40 V 2022 ww 1 Figure 3.61 For Prob. 3.12.
Please find Vo using nodal analysis
2 kly + 1 k 1 k2 1 kn 1 kn 12 V O +
Chapter 8, Problem 8.073 Use nodal analysis to find Vo in the circuit in the figure below. 6/00 V 1 0 10 0 1Ω 2/0° A (a) Find the real part of Vo. (b) Find the imaginary part of V Vi Click if you would like to Show work for this question: Open-Show-work
(b) Assume that you have been asked to find V. in the circuit of Figure Q2(b). (1) Between mesh and nodal analysis technique, which method would you recommend? Explain why. [2 marks] (11) Use your recommended method of analysis to find Vo. [13 marks] -j5 Ω j2 Ω 8 Ω, 1220° Α 5Ω ΥΥΥ j8 Ω +V- + 15Z90° V -j2 Ω 1 10240° A Figure Q2(b)