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this is part of the solution of a previous exam and there are some things i don't understand about it

1- why did he assume the value of current at the 12 Ohm resistor to be 0.5A without any calculations?

2- how did he assume the polarity of Vx in the bottom?

3-if we didn't assume the current of the 12 Ohm resistor can we figure it out by doing a KVL loop between the 12 Ohm and the 6 Ohm resistors?

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1 Answer 1

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1: Because the \$12\Omega\$ and the \$6\Omega\$ resistors are in parallel, and the current through the \$6\Omega\$ resistor is \$1A\$, then necessarily, the current through the \$12\Omega\$ resistor must be \$0.5A\$. Think about why that is.

2: The polarity of \$V_x\$ is completely arbitrary. Swapping the signs around would just put a negative in front of anywhere that it appears in the resulting equations. The "natural" polarity (due to the passive sign convention), however, is to mark it as done in the problem, since the current \$5i_x - 1.5\$ is drawn entering the resistor to the right.

3: Yes, and in fact, that's exactly where the answer to #1 comes from.

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