I've tried using KVL and KCL but nothing I do is resulting in an answer. I am struggling with how to treat the voltage drop and the two resistors that are in parallel with each other. I think I may be trying to put a ground node in the wrong place.

Any help to get started would be greatly appreciated.

The question says to use KVL and KCL to solve for v.

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  • \$\begingroup\$ I see two resistors in series - nothing in parallel with anything else. \$\endgroup\$ – Peter Bennett Apr 1 '15 at 1:15

The system is complete within itself. You don't need to reference ground. KCL means that only \$i_x\$ flows around the loop. Voltage drop in resistors is \$i_xR\$ so each resistor drops \$2i_x\$ volts. So using KVL; $$12V- 2i_x -2i_x -6i_x -2i_x = 0V$$ $$ Ix = 12/12 = 1\ Amp.$$


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