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WorksheetsNodal
Total questions: 10
Worksheet time: 5mins
Derive the equation for node V1
30∠−90°=(0.5+j0.4)V1−j0.4V2
30∠90°=(0.5+j0.4)V1−j0.4V2
30∠−90°=(0.5−j0.4)V1−j0.4V2
30∠−90°=(0.5+j0.4)V1+j0.4V2
which is the correct equation for node V2?
0=−(0.75+j0.4)V1+(0.25+j0.15)V2
0=−(0.75−j0.4)V1+(0.25+j0.15)V2
0=−(0.25+j0.15)V1+(0.75+j0.4)V2
0=−(0.25−j0.15)V1+(0.75+j0.4)V2
Should we used super-node in this circuit?
Yes
No
what is the expression for Vx?
V1
V1-V2
V2
V2-V1
what is the equation for super-node?
36=(1−j2)V1+j4V2
36=j4V1+(1+j2)V2
−36=(1−j2)V1+j4V2
36=j4V1+(1−j2)V2
which equation is correct for KVL at super-node?
V1=V2−10∠45°
10∠45°=V2+V1
V1=V2+10∠45°
V2=V1+10∠45°
what is the equation for super-node?
15=(0.25−j0.25)V1+(0.5+j)V2
3.75=(0.25−j0.25)V1+(0.5+j)V2
3.75=(5+j0.25)V1+(0.5+j)V2
3.75=(0.25+j0.25)V2+(0.5+j)V1
Do we need to use super-node to solve this circuit?
Yes
No
which conversion of time domain to frequency domain is correct?
ω=4k srad, ZL=j1kΩ, ZC=−j125Ω
ω=4k srad, ZL=−j1kΩ, ZC=j125Ω
ω=400 srad, ZL=j100Ω, ZC=−j12.5Ω
ω=4k srad, ZL=j1kΩ, ZC=j125Ω
derive the equation at the node 1
io=j100V1−−j125(V1−10io)
io=jkV1+j125(V1+10io)
io=jkV1+−j125(V1−10io)
io=jk(V1−10io)+−j125V1
