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1. Since there are only two nodes, it seems appropriate to apply KCL.
2. Alternatively you can transform the 6A current source into voltage source and then apply KVL
The attachment shows the solution using the first method (ie. by applying KCL)
To convert a current source in parallel with a resistor to its equivalent voltage source we find its Thevinin's equivalent.
This simply results in a voltage source in series with the same resistor
The voltage is given simply by IR
Yes, use the node method. There is only one node. Once you find the voltage of the node, then you can find all the currents in the branches.
Current going into the node equals the current leaving the node (KCL). Six amps are entering the node, V/2 +(1/4)*(V/2)+V/8 amps are leaving the node. Can you find V now?
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