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Any help would be appreciated.
Write your mesh current equations for each of the meshes, for example, \[i_2 = -20 + R(i_2-i_1) + 16 = 0 \] Plugging the values you are given will enable you to solve for your voltage drop on R. Similarly, you know the voltages and currents in each of the voltage sources, so you can calculate either P=-vi or P=vi, depending on which direction your current is flowing. Your current source current, since direction is indicated, is \[i_3 - i_1\] Write down everything you know, write your mesh equations, fill in the blanks, solve for unknowns.

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Other answers:

by looking at the units you have current equal to voltage, plug and play is not available from this form. Perhaps the following:\[i_{2} = \frac{20V-16V}{R}\]
@Mathmuse You cannot to this type of relationship because i2 is a mesh, you also cannot do a simple subtraction of the voltage sources as it does not take into account the voltage drop on the resistor or the impact on that loop by the rest of the circuit.
ahh yes...i didn't take into accountt i1.
If you want to completely learn, mesh analysis visit

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