2
\$\begingroup\$

enter image description here

How can I approach this type of problem? Which different methods can be used here? KVL, supernode, mesh-analysis, source transform.

I tried using KCL on a node assuming its voltage V1:

2=(V1-10)/1 + (V1-4Vo)/4
Solving we get eqn :
5V1-4Vo=48  ...(1)

How can I form the second equation?

Assuming loop currents i1 and i2 (clockwise),
so i1=-2 
Vo=(i2+2)*1 ..(total current is i2+2 isn't it?)
second loop eqn is:
-10-Vo+4i2+4Vo=0
3Vo+4i2=10
3(i2+2)+4i2=10
3i2+6+4i2=10
i2=4/7 Amp (Probably wrong! Why?)
Vo=(4/7)+2= 2.57
4Vo=10.28
Power=(10.28)^2 * (4/7)=Wrong
the 4 answer options are +/- 1152 or +/- 1200
\$\endgroup\$
4
  • 1
    \$\begingroup\$ The correct answer is 1152 watts BTW. 16 amps flows through the 4 ohm resistor. \$\endgroup\$
    – Andy aka
    Apr 9 at 13:08
  • \$\begingroup\$ P=I^2*R so 16*16*4= 1024 .How you got 1152? can you elaborate the mistake in my approach \$\endgroup\$
    – Adyy
    Apr 9 at 13:10
  • \$\begingroup\$ I just simulated it for convenience. Modern sims give very exact results. If you simulate you can begin to see the relationships and double check your math. \$\endgroup\$
    – Andy aka
    Apr 9 at 13:11
  • \$\begingroup\$ Without simulations what to do to find i in right loop still can't find mistakes, is Vo=-2 if so then 4Vo is -8 and to get 1152 Watts i have to have 144 amps current in second loop which seems too large? \$\endgroup\$
    – Adyy
    Apr 9 at 13:36

2 Answers 2

Reset to default
2
\$\begingroup\$

How to approach this type of problem? (KVL, supernode, mesh-analysis, source transform)?

These are my KVL equations: -

enter image description here

So, from the first equation: -

$$3\cdot V_0 = 18 - 4\cdot I_1$$

Substitute the 2nd equation into the first equation: -

$$3\cdot (-I_1) = 18 - 4\cdot I_1$$

Hence \$\hspace{1cm}I_1 = 18 \$ amps \$\hspace{1cm}\$and,\$\hspace{1cm}\$ \$V_0 = -18\$ volts $$$$ This of course means that the voltage magnitude across the dependent source is 72 volts and, the current flowing into it is 16 amps. This makes it dissipate 1152 watts.

\$\endgroup\$
1
\$\begingroup\$

Your (V1-10) happens also be =Vo. Use it in the rightmost branch current in the KCL equation.

\$\endgroup\$
4
  • \$\begingroup\$ 5(Vo +10)-4Vo=48 which means Vo=-2 putting that in KCL loop -10+i+2+4i+4(-2)=0 ; i =16/5 and power=Vxi , 8 * 16/5 == 25.6, but the answer is either +/- 1152 or +/- 1200 Watts. What is the mistake here ? \$\endgroup\$
    – Adyy
    Apr 9 at 13:03
  • 1
    \$\begingroup\$ Do as I said, get V1=-8 volts. Then calculate Vo=-18 volts. The right side branch current is 16A upwards. 4Vo is -72 volts, The actual plus pole is ithe bottom side.. The dependent source sinks 1152 watts. \$\endgroup\$
    – H. Gebhard
    Apr 9 at 13:45
  • \$\begingroup\$ Yes Got it, V1=-8 Solves everything i was getting V1=+8. I mean when we apply KCL don't we assume the node to be at higher potential than others , So (V1-4Vo)/4 + (V1-10)/1 = 2 which gives V1=+8? \$\endgroup\$
    – Adyy
    Apr 9 at 14:17
  • \$\begingroup\$ The right KCL starting eq, is 2+(V1-10)/1+(V1-4Vo)/4=0 ; that means zero total current out of the mid top node. You have already a wrong starting eq. \$\endgroup\$
    – H. Gebhard
    Apr 9 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.