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One of my friends sent me this question but I have never seen anything like that, anyone knows what is it or how can I solve it? Question asks: what's the value of alpha and calculate the power of current source using alpha?

Also what's that thing in the middle?

enter image description here

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    \$\begingroup\$ The ideal source in the center is a current dependent voltage source. The voltage produced by it is proportional to the current i-delta. \$\endgroup\$ Commented Oct 16, 2019 at 20:46
  • \$\begingroup\$ So delta is 400 ? (6v/15ma) and power is 90mW ? Also if I understand it right right part is some kind of current meter ? \$\endgroup\$
    – Mordecai
    Commented Oct 16, 2019 at 20:48
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    \$\begingroup\$ The question title is among the worst I've seen, but I'm stumped as to what it should be. \$\endgroup\$
    – JYelton
    Commented Oct 16, 2019 at 20:54
  • \$\begingroup\$ @JYelton Yeah sorry about that , language differences making it really problematic \$\endgroup\$
    – Mordecai
    Commented Oct 16, 2019 at 20:56
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    \$\begingroup\$ Revisiting this, I have changed the title to be more applicable. \$\endgroup\$
    – JYelton
    Commented Oct 18, 2019 at 21:09

1 Answer 1

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The thing in the middle is a dependent voltage source. It is a voltage source that has a variable voltage, the value of which depends on the given formula.

Now on to solving this puzzle... (it really is much more of a puzzle than a circuit)

The 6 V independent source forces the voltage between the two nodes to be 6 V.

This means that the voltage across the dependent source must also be 6 V. As the +/- signs are reversed, this means that \$\alpha i_\Delta = -6\text{ V}\$. But we know that \$i_\Delta = -15\text{ mA}\$ looking at the current through the current source.

Plugging in \$i_\Delta\$ into the first equation, we get

$$\alpha\cdot(-15\text{ mA}) = -6\text{ V}$$

So

$$\alpha = 400\text{ Ω}$$

You can actually calculate the power of the current source without doing any of this, since \$P = IV\$ and you know both \$I\$ and \$V\$. Just multiply 6 V times 15 mA and you get 90 mW.

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  • \$\begingroup\$ Okay thanks for answer , I have never seen anything like that before . Also I want to be clear , current source's power is 90mW and right part is some kind of current meter ? \$\endgroup\$
    – Mordecai
    Commented Oct 16, 2019 at 20:59
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    \$\begingroup\$ @Mordecai - All three of the circuit elements here are sources. Circular ones are independent sources, meaning the value is constant, and diamond ones are dependent sources, meaning the value changes based on the given formula. If the source has + - in it, it is a voltage source, whereas if it has an arrow it is a current source. In summary, the right part is an independent current source. \$\endgroup\$
    – Justin
    Commented Oct 17, 2019 at 12:31
  • \$\begingroup\$ The part of this circuit that could be called a current meter is the little blue arrow on the top next to \$i_\Delta\$. \$\endgroup\$
    – Justin
    Commented Oct 17, 2019 at 12:32
  • \$\begingroup\$ I researched the topic but whole idea seem like "imaginary" or "ideal" . You can't use this in real life circuits , can you ? \$\endgroup\$
    – Mordecai
    Commented Oct 18, 2019 at 21:21
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    \$\begingroup\$ @Mordecai, you can't put one (ideal) voltage source in parallel with another in a real circuit. But there are lots of devices (transistors, for example) that are modeled with dependent sources. \$\endgroup\$
    – The Photon
    Commented Oct 18, 2019 at 21:36

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