As in the previous examples, we label the nodes, define the directions of the currents, and define the voltages using the sign convention.

We note that the 2 Ωresistor is short-circuited and can be omitted in the analysis. Using
KVL and KCL, we obtain
• KVL(1 → 2 → 1): 3i_{x} + 24 = 0 \longrightarrow i_{x} = −8A,
• KCL(2): i_{x} + 1 + 6 + i_{y} = 0 \longrightarrow i_{y} = 1A,
• KVL(1 → 2 → 3 → 1): −24 − 1i_{y} + v_{s} = 0 \longrightarrow v_{s} = 25V.

As briefly discussed before, it is generally suggested to avoid using KCL at a node with a connection to a voltage source. However, in this case, the current across the voltage source v_{s} must be found in order to find the voltage value.Therefore, we apply KCL at node 2.