KCL and KVL make a circuit solvable

Conservation becomes an equation

Kirchhoff's current law (KCL) says the algebraic sum of currents at a node is zero. In a lumped circuit model, charge does not accumulate indefinitely at a wire junction. Pick currents entering as positive, or currents leaving as positive, and keep that choice. If 2 mA enters and two branches carry 0.7 mA and an unknown outward current, the unknown is 1.3 mA.

Kirchhoff's voltage law (KVL) says the algebraic sum of voltage changes around a closed loop is zero. It applies to the lumped model used here, where any induced voltage is represented by a circuit element. Walking from a source's negative terminal to its positive terminal is a rise; walking along a resistor's current is a drop.

Worked divider

Connect a 5 V source, a 2 kΩ upper resistor, and a 3 kΩ lower resistor in series. Let the current be clockwise. KVL gives $5-I(2000)-I(3000)=0$, so $I=1$ mA. Taking the bottom node as ground, the middle node is 3 V. The upper drop is 2 V and the lower drop is 3 V; their sum equals the source voltage.

5 V --- 2 kohm --- out --- 3 kohm --- 0 V
              same series current: 1 mA

At out, KCL gives upper current minus lower current equal to zero. This statement changes as soon as another branch is connected. A divider formula is therefore a consequence of the topology, rather than a shortcut that survives any load.

Check the answer twice

The 2 kΩ resistor absorbs 2 mW, the 3 kΩ resistor absorbs 3 mW, and the source supplies 5 mW. Both the loop-voltage sum and power sum close. If your computed out voltage exceeds 5 V in this passive DC network, inspect signs or topology. An equation producing a negative branch current is acceptable: it says the actual direction is opposite your chosen arrow. Keep the sign instead of silently reversing only part of the calculation.