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Mr_Good_Cat_67
 2 years ago
Best ResponseYou've already chosen the best response.0Any help would be appreciated.

eSpeX
 2 years ago
Best ResponseYou've already chosen the best response.0Write your mesh current equations for each of the meshes, for example, \[i_2 = 20 + R(i_2i_1) + 16 = 0 \] 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.

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.0by looking at the units you have current equal to voltage, plug and play is not available from this form. Perhaps the following:\[i_{2} = \frac{20V16V}{R}\]

eSpeX
 2 years ago
Best ResponseYou've already chosen the best response.0@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.

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.0ahh yes...i didn't take into accountt i1.

knowelectronics
 2 years ago
Best ResponseYou've already chosen the best response.0If you want to completely learn, mesh analysis visit www.knowelectronics.org
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