Circuit Solver

Every DC network is solved with Ohm's law plus Kirchhoff's two conservation laws:

$\\sum_{\\text{node}} I = 0 \\qquad \\sum_{\\text{loop}} V = 0$

• Kirchhoff's current law (KCL): charge cannot pile up at a junction, so the currents in equal the currents out, in amperes (A).
• Kirchhoff's voltage law (KVL): going once around any closed loop returns you to the same potential, so the rises and drops in volts (V) cancel.

With $n$ unknown node voltages you write $n$ KCL equations and solve the linear system. Two nodes give two equations, and the arithmetic is ordinary simultaneous-equation work.

The operating assumption: steady-state DC with ideal, lumped components — wires of zero resistance, sources of zero internal resistance unless one is stated, and no capacitance or inductance. Add reactance and you must work with impedance $Z$ and phasors instead.

The mistake people make: assigning current directions and then refusing to accept a negative answer. A negative current simply means the real flow is opposite to the arrow you drew, and the number is still correct.

Most textbook circuits collapse without any simultaneous equations. Work from the far end back to the source:

$R_s = R_1 + R_2 + \\cdots \\qquad \\frac{1}{R_p} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\cdots$

Find the total resistance, get the supply current from $I = V/R$, then expand back out, using $V = IR$ at each step to recover the individual voltages and currents.

Two shortcuts save most of the work:

$V_{\\text{out}} = V_{\\text{in}}\\frac{R_2}{R_1 + R_2} \\qquad I_1 = I_{\\text{total}}\\frac{R_2}{R_1 + R_2}$

the voltage divider and the two-branch current divider. Note the current divider takes the opposite resistance on top: more current goes through the smaller resistor.

When reduction fails: a bridge network has no two elements purely in series or purely in parallel. Then you need node voltage analysis, mesh analysis, or a delta-wye transformation. Recognising that early saves a wrong answer.

• Calling elements parallel when they only look parallel — two components are in parallel only if both ends share a node.
• Forgetting the final reciprocal in $1/R_p$.
• Using the source voltage for one series element — only the drop across that element belongs in $V = IR$.
• Ignoring the load on a divider — connecting a load across the lower resistor changes the output, often dramatically, as the third example shows.
• Losing signs in KVL — traverse the loop in one consistent direction and treat a drop as negative throughout.
• Assuming the same current everywhere — that is only true within a single series path; at a node the current splits.
• Neglecting source internal resistance — a battery under a heavy load delivers less than its open-circuit voltage.