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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

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    \$\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

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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.

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

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  • \$\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
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    \$\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

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