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So I'm having some trouble getting to the answer of this question (which I have been told is C), and I've been trying to solve it via mesh analysis, but the final answer I get seems to be off completely.

I think that the reason my answer is off might be because of my equation for the second mesh since I'm getting a weird fraction for the current, but I'm not sure how I would change it :/

So I'm having some trouble getting to the answer of this question (which I have been told is C), and I've been trying to solve it via mesh analysis, but the final answer I get seems to be off completely.

I think that the reason my answer is off might be because of my equation for the second mesh since I'm getting a weird fraction for the current, but I'm not sure how I would change it :/

Any help would be much appreciated.

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  • \$\begingroup\$ Hint: there's no way to write the voltage across the CCCS in terms of mesh currents. But the CCCS does allow you to write an equation relating I1, I2, and I3. \$\endgroup\$ – The Photon Apr 19 '18 at 18:25
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You can take the help of Supermesh.

By the way, how about taking a closer look at the current in each branch?

Mesh equations

Hope this helps!

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  • \$\begingroup\$ I'm a bit confused on how the current through the 22.5 ohm resistor is works. Is the V/22.5 coming from the dependent current source? Is it a part of it? \$\endgroup\$ – seph22 Apr 19 '18 at 20:09
  • \$\begingroup\$ As we have already defined the polarity of voltage across 22.5 ohm resistor, based on that assumption we consider the direction of current to flow from a point of higher potential to a lower one (even in our calculations). P.S. After solving the unknowns, a close observation will tell you that the 50V battery is getting charged (power is fed into it), and all the resistors contribute to power loss. Hence, all the power is supplied by the CCCS alone. \$\endgroup\$ – Pranabendra Apr 20 '18 at 3:41
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It appears that Pranabendra beat me to providing a worked answer, but I thought I would explain what is wrong with your original attempt.

Mesh analysis is based on Kirchoff's voltage law, which states that if we consider the voltages around any closed loop in a circuit we have $$ V_1 + V_2 + \ldots + V_n = 0$$ so that all the potential differences add to zero. The idea of mesh analysis is that if we can write the voltage across each component as either a constant or a simple function of the current through that component, then KVL allows us to write down an equation for the loop currents: $$ V_1(i_1,\ldots,i_n) + V_2(i_1,\ldots,i_n) + \ldots + V_n(i_1,\ldots,i_n) = 0$$ This is easy for voltage sources and resistors, but for current sources it's not possible since the voltage across a current source depends upon the circuit it is connected to.

In your circuit there is an unknown potential difference across the current source, so your original equation for M2 was incorrect as you ignored this voltage. The way to remedy this is to write an equation for a loop that doesn't include the current source (e.g. the supermesh used by Pranabendra).

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Kirchhoff's Voltage Law (KVL) says the the sum of voltages around closed loop is zero. This is the basis for Mesh Analysis.

In your equation for the M1 mesh (which is correct), you have the voltages across the source and two resistors summing to zero. But the equations for M2 and M3 are missing the voltage across the dependent current source. This is why you're not able to get the correct answer.

It's more convenient to work this problem using Nodal Analysis, which is based on KCL. There are only two node voltages to solve for, and expressions for the currents are straightforward.

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  • \$\begingroup\$ I did initially try to use nodal analysts, but had a '3I' in one of the equations, so i ended up with two equations with three unknowns, being V1, V2 and 3I. Unless I was doing the equation for across the current source wrong as well which might be the case... \$\endgroup\$ – seph22 Apr 19 '18 at 19:37
  • \$\begingroup\$ You need to substitute for I = (V1 - 50) / 5 ohms. \$\endgroup\$ – user28910 Apr 20 '18 at 15:04

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