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Nodal

Total questions: 10

Worksheet time: 5mins

Name
Class
Date
1.

Derive the equation for node V1

a)

 3090°=(0.5+j0.4)V1j0.4V230\angle-90\degree=\left(0.5+j0.4\right)V_1-j0.4V_2  

b)

 3090°=(0.5+j0.4)V1j0.4V230\angle90\degree=\left(0.5+j0.4\right)V_1-j0.4V_2  

c)

 3090°=(0.5j0.4)V1j0.4V230\angle-90\degree=\left(0.5-j0.4\right)V_1-j0.4V_2  

d)

 3090°=(0.5+j0.4)V1+j0.4V230\angle-90\degree=\left(0.5+j0.4\right)V_1+j0.4V_2  

2.

which is the correct equation for node V2?

a)

0=(0.75+j0.4)V1+(0.25+j0.15)V20=-\left(0.75+j0.4\right)V_1+\left(0.25+j0.15\right)V_2

b)

0=(0.75j0.4)V1+(0.25+j0.15)V20=-\left(0.75-j0.4\right)V_1+\left(0.25+j0.15\right)V_2

c)

0=(0.25+j0.15)V1+(0.75+j0.4)V20=-\left(0.25+j0.15\right)V_1+\left(0.75+j0.4\right)V_2

d)

0=(0.25j0.15)V1+(0.75+j0.4)V20=-\left(0.25-j0.15\right)V_1+\left(0.75+j0.4\right)V_2

3.

Should we used super-node in this circuit?

a)

Yes

b)

No

4.

what is the expression for Vx?

a)

V1

b)

V1-V2

c)

V2

d)

V2-V1

5.

what is the equation for super-node?

a)

36=(1j2)V1+j4V236=\left(1-j2\right)V_1+j4V_2

b)

36=j4V1+(1+j2)V236=j4V_1+\left(1+j2\right)V_2

c)

36=(1j2)V1+j4V2-36=\left(1-j2\right)V_1+j4V_2

d)

36=j4V1+(1j2)V236=j4V_1+\left(1-j2\right)V_2

6.

which equation is correct for KVL at super-node?

a)

V1=V21045°V_1=V_2-10\angle45\degree

b)

1045°=V2+V110\angle45\degree=V_2+V_1

c)

V1=V2+1045°V_1=V_2+10\angle45\degree

d)

V2=V1+1045°V_2=V_1+10\angle45\degree

7.

what is the equation for super-node?

a)

15=(0.25j0.25)V1+(0.5+j)V215=\left(0.25-j0.25\right)V_1+\left(0.5+j\right)V_2

b)

3.75=(0.25j0.25)V1+(0.5+j)V23.75=\left(0.25-j0.25\right)V_1+\left(0.5+j\right)V_2

c)

3.75=(5+j0.25)V1+(0.5+j)V23.75=\left(5+j0.25\right)V_1+\left(0.5+j\right)V_2

d)

3.75=(0.25+j0.25)V2+(0.5+j)V13.75=\left(0.25+j0.25\right)V_2+\left(0.5+j\right)V_1

8.

Do we need to use super-node to solve this circuit?

a)

Yes

b)

No

9.

which conversion of time domain to frequency domain is correct?

a)

ω=4k rads, ZL=j1kΩ, ZC=j125Ω\omega=4k\ \frac{rad}{s},\ Z_L=j1k\Omega,\ Z_C=-j125\Omega

b)

ω=4k rads, ZL=j1kΩ, ZC=j125Ω\omega=4k\ \frac{rad}{s},\ Z_L=-j1k\Omega,\ Z_C=j125\Omega

c)

ω=400 rads, ZL=j100Ω, ZC=j12.5Ω\omega=400\ \frac{rad}{s},\ Z_L=j100\Omega,\ Z_C=-j12.5\Omega

d)

ω=4k rads, ZL=j1kΩ, ZC=j125Ω\omega=4k\ \frac{rad}{s},\ Z_L=j1k\Omega,\ Z_C=j125\Omega

10.

derive the equation at the node 1

a)

io=V1j100(V110io)j125i_o=\frac{V_1}{j100}-\frac{\left(V_1-10i_o\right)}{-j125}

b)

io=V1jk+(V1+10io)j125i_o=\frac{V_1}{jk}+\frac{\left(V_1+10i_o\right)}{j125}

c)

io=V1jk+(V110io)j125i_o=\frac{V_1}{jk}+\frac{\left(V_1-10i_o\right)}{-j125}

d)

io=(V110io)jk+V1j125i_o=\frac{\left(V_1-10i_o\right)}{jk}+\frac{V_1}{-j125}