Kirchhoff’s Laws and Network Theorems

Tip

Exam Focus

Very high yield: state KCL/KVL with their conservation basis; solve a two-mesh circuit (KVL) and a node (KCL); superposition (one source at a time, $V\!\to$short, $I\!\to$open, never for power); Thevenin/Norton equivalents and the source transformation $V_{Th}=I_NR_N$; and maximum power transfer $R_L=R_{Th}$ with $\eta=50\%$. Star–delta (asked NTC 2082) is in the next chapter.

Kirchhoff’s Laws

Note

Law — Kirchhoff’s current and voltage laws

KCL (charge conservation): the algebraic sum of currents at a node is zero,

$$\sum I=0\quad\Longleftrightarrow\quad \sum I_{\text{in}}=\sum I_{\text{out}}.$$

KVL (energy conservation): the algebraic sum of voltage rises and drops around any closed loop is zero,

$$\sum_{\text{loop}} V=0.$$

KCL balances currents at a node; KVL balances the source rise against the resistor drops around a loop.

Mesh and Nodal Analysis

Mesh analysis assigns a clockwise current to each window of a planar circuit and writes KVL per mesh. Nodal analysis picks a reference node and writes KCL at each remaining node in terms of node voltages.

Two meshes sharing $R_3$: the shared branch carries $I_1-I_2$.

Caution

Key Formula — Two-mesh KVL / nodal equations

$$(R_1+R_3)I_1-R_3I_2=V_1,\qquad -R_3I_1+(R_2+R_3)I_2=0.$$

For a floating current source between meshes use a supermesh; for a floating voltage source between nodes use a supernode.

[!NOTE]

Example — Mesh solution

With $R_1=2\,\Omega$, $R_2=4\,\Omega$, $R_3=2\,\Omega$, $V_1=10\,V$: $4I_1-2I_2=10$ and $-2I_1+6I_2=0$ give $I_1=3\,A$, $I_2=1\,A$; the shared branch carries $I_1-I_2=2\,A$.

Superposition Theorem

Note

Key Point

In a linear circuit the response equals the algebraic sum of the responses to each independent source acting alone, with the others killed: voltage source $\to$ short, current source $\to$ open. Dependent sources are never killed. It applies to voltage/current but not directly to power ($P\propto I^2$ is non-linear).

Thevenin, Norton and Maximum Power Transfer

Any linear two-terminal network reduces to a Thevenin source ($V_{Th}$ in series with $R_{Th}$) or the equivalent Norton source ($I_N$ in parallel with $R_N$).

Thevenin and Norton equivalents of a linear network driving a load $R_L$.

Caution

Key Formula — Thevenin/Norton and maximum power transfer

$$V_{Th}=V_{oc},\quad R_{Th}=R_N,\quad I_N=I_{sc}=\frac{V_{Th}}{R_{Th}};\qquad \boxed{R_L=R_{Th}}\Rightarrow P_{\max}=\frac{V_{Th}^2}{4R_{Th}},\ \eta=50\%.$$

[!NOTE]

Example — Thevenin equivalent

A 12 V source with a 4 $\Omega$ series resistor feeds a 6 $\Omega$ branch to ground. Then $V_{Th}=12\cdot\frac{6}{4+6}=7.2\,V$ and $R_{Th}=4\parallel6=2.4\,\Omega$. Matched load $R_L=2.4\,\Omega$ draws $I_L=1.5\,A$ and absorbs $P_{\max}=7.2^2/(4\cdot2.4)=5.4\,W$.