Answer: B. 45. What is correct equation for node B (Fig.7), if we apply node voltage analysis?
45. What is correct equation for node B (Fig.7), if we apply node voltage analysis? (G=1/R) A.– 2GvA- 0.5GvB+3.5GvC=0 B. 2GvA- 0.5GvB+3.5GvC=0 C. 0.5GvA+3GvB-0.5GvC =0 D. – 0.5GvA+3GvB-0.5GvC=0
Solution: VB(0.5G+2G+0.5G) –VA(0.5G) –VC(0.5G)=0 –0.5GVA+3GVB–0.5GVC=0 Answer: D.
46. Formulate mesh-current equation for mesh A for the circuit in Fig.8.
C. (R3+R2)iC-R3iC- R2iA= vS1 D. (R3+R2)iC-R3iC- R2iA= -vS1 E. (R1+R2)iA-R2iC= - vS2
Solution: iA(R1+R2) – iCR2= –VS2 Fig.8 Answer: E.
47. To apply superposition principle we must ______ in turn all sources, except one. When the current source is _________, we replace it by _________. A. “turn on”, turned off, short circuit. B. “turn off”, turned on, open circuit. C. “turn off”, turned on, short circuit. D. “turn off”, turned off, open circuit. E. “turn off”, turned off, short circuit.
Solution: To apply superposition principle we must “turn off” all sources in turn. When the current source is turned off, we replace it be open circuit. Answer: D.
48. Ideal voltage source is described as v=____ and i=____, but ideal current source – i=_____, v=_____. A. vS, any value, iS, any value B. any value, iS, any value, vS C. vS, iS, any value, any value D. any value, any value, iS, vS E. all answers are wrong
Solution: Ideal voltage source is described as v= vS and i= any value, but ideal current source – i= iS, v= any value. Answer: A.
49. Find the correct formula for p. p = A. vi B. dw/dq C.dq/dt D. dq/dv E. di/dt Solution: p=dw/dt=vi. Answer: A.
50. The parameters of the Thevenin and Norton equivalent circuits at a given interface can be found as A. vT=vSC iN=iOC RN=RT=vOC/iSC B. vT=vOC iN=iSC RN=RT=vOC/iOC C. vT=vSC iN=iOC RN=RT=vSC/iSC D. vT=vSC iN=iOC RN=RT=vOC/iOC E. vT=vOC iN=iSC RN=RT=vOC/iSC
Solution: vT=vOC iN=iSC RN=RT=vOC/iSC
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E. 2.5GvA-0.5GvB -2GvC=iS Fig. 7
A. (R3+R4)iB-R3iC= vS2 B. (R1+R2)iA-R2iC= vS2
