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My attempt at the problem starts with,

-I assumed that the value of the independent current source which is \$10 mA\$ is the same throughout all the elements since it is a series circuit (Please tell me if this is right, because I based all my #s on this)

-Knowing this, I calculated \$V2\$ using \$V=I R\$

\$V2=(10 mA)(1 k\Omega)= 100 V\$

-I calculated P2 & P3 using \$P=V I\$,

\$P2=(100v)(10mA)=1 W\$

\$P3=(10v)(10mA)=0.1 W\$

Now I'm only left to find \$V1\$ & \$P1\$

By using passive convention, in the end, the sum of power is going to equal zero

-so I assumed that,

\$(-P1)=P2 + P3\$

\$P1= -1.1 W\$

So what's left is \$V1\$, and knowing that \$P1\$ is a negative value, I thought that somehow, the direction or polarity of the \$I1\$ or \$V1\$ needs to be in the negative direction... but I wasn't sure how to draw in a way that would satisfy passive convention. I got stuck from here.

Somebody please help me, and correct me if I made false assumptions. Thank you in advance.

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  • \$\begingroup\$ There are two paths from ground (assumed lower conductor) to the top junction of the current source. One is via the current source, and we don't know the voltage across this yet. The other is up through the voltage source by 10V and up by a further 100V through the resistor. Hence... \$\endgroup\$
    – Chu
    Commented Sep 7, 2018 at 7:45

1 Answer 1

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...is 10mA is the same throughout all the elements That is correct.

V2=(10mA)(1k)= 100V You mean 10 mA * 10 k = 100 V answer is correct though

The power of V3, think about that for a moment.

If you take a voltage source like V3 and it delivers power to a load like a resistor, what direction is the current flowing? What direction is the current flowing in this circuit? Is it the same direction or the opposite direction.

So does V3 deliver power (like when there is only a load resistor) or does it absorb power?

Why do you assume that the sum of all powers must be zero? Does "passive convention" apply here? I do see sources that can deliver and absorb power so..

How about simply calculating the voltage across source I1, that's simple. You already know the current as well so the power can be easily calculated.

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