# Voltage drop across a series with a source and a resistor

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.

• I see two resistors in series - nothing in parallel with anything else. – 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; $$12V- 2i_x -2i_x -6i_x -2i_x = 0V$$ $$Ix = 12/12 = 1\ Amp.$$