Kirchhoff's Laws

Terminologies for electric circuits:

• node: a point where three or more current-carrying elements (branches) are connected;
• branch: a path containing one or more elements in series and connecting two nodes;
• loop: a closed path progressing from node to node and returning to the starting node.

Conventionally, constant variables are represented in upper-case letters (e.g., DC voltage $V$ and current $I$, etc.), and time-varying variables are treated as time functions and represented in lower-case letters (e.g., AC voltage $v(t)$ and current $i(t)$).

Also note that in a circuit diagram, the direction of a current and the polarity of a voltage source can be assumed arbitrarily. To determined the actual direction and polarity, the sign of the values also should be considered. For example, a current labeled in left-to-right direction with a negative value is actually flowing right-to-left.

Kirchhoff's Laws

Based on the principle of conservation of electric charge, the Kirchoff current Law (KCL) states that:

The algebraic sum of the currents into a node at any instant is zero.

$$\sum_k I_k=0$$

Based on the principle of conservation of energy, the Kirchoff voltage Law (KVL) states that:

The algebraic sum of the voltages around a loop at any instant is zero.

$$\sum_k V_k=0$$

Assume currents going into the node are positive and those leaving the node negative, KCL states: $4+5-3-4-2=0$.

Assume the current goes around a clockwise loop, KVL states: $-12+3+4+5=0$; alternatively, if assume the current goes around a counter clockwise loop, we have $12-5-4-3=0$;

Example: Given circuit below, find $V_2$, $V_0$, $I_2$, $R_1$ and $R_2$.

According to Ohm's law, we have $I_2=3/2=1.5A$.

Apply KVL to the loop on the right to get:

$$V_2-5+3=0,\;\;\;\;\;\; V_2=2V$$

According to Ohm's law, we have $R_2=V_2/I_2=2V/1.5A=1.33\Omega$.

Apply KCL to the middle node on top to get:

$$2-I_1-I_2=2-I_1-1.5=0,\;\;\;\;\;I_1=0.5A$$

Again by Ohm's law, we get $R_1=5V/0.5A=10\Omega$.

Apply KVL to the loop on the left to get:

$$3\times 2 +5-V_0=0,\;\;\;\;\;\;V_0=11V$$

Voltage Divider:

According to Ohm's law, the voltage $V_k$ across the kth resistor $R_k$ can be found to be:

$$V_k=IR_k=\frac{V}{R_s}R_k=V\frac{R_k}{R_1+R_2+\cdots+R_n}$$

In particular if $n=2$, we have

$$V_1=V\frac{R_1}{R_1+R_2},\;\;\;\;\;\;\;\;V_2=V\frac{R_2}{R_1+R_2}$$

Resistors in series: According to KVL, the sum of voltages across the resistors is equal to the input voltage:

$$V=V_1+V_2+\cdots+V_n=I(R_1+R_2+\cdots+R_n)=IR_s$$

i.e.,

$$R_s=\frac{V}{I}=R_1+R_2+\cdots+R_n$$

Resistors in parallel: According to KCL, the sum of currents through the resistors is equal to the input current:

$$I=I_1+I_2+\cdots+I_n=\frac{V}{R_1}+\frac{V}{R_2}+\cdots+\frac... ...}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}\right)=\frac{V}{R_p}$$

i.e.,

$$\frac{I}{V}=\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n} =G_p=G_1+G_2+\cdots+G_n$$

or

$$R_p=\frac{1}{1/R_1+1/R_2+\cdots+1/R_n}=R_1\vert\vert R_2\vert\vert\cdots\vert\vert R_n$$

In particular, when $n=2$,

$$R_p=R_1\vert\vert R_2=\frac{1}{1/R_1+1/R_2}=\frac{R_1\;R_2}{R_1+R_2}$$

Current Divider:

According to Ohm's law, the current $I_k$ through the kth resistor $R_k$ can be found to be:

$$I_k=\frac{V}{R_k}=\frac{IR_p}{R_k}=I\frac{1/R_k}{1/R_1+1/R_2+\cdots+1/R_n} =I\frac{G_k}{G_1+G_2+\cdots+G_n}$$

In particular if $n=2$, we have

$$I_1=I\frac{G_1}{G_1+G_2}=I\frac{R_2}{R_1+R_2} ,\;\;\;\;\;\;\;\;I_2=I\frac{G_2}{G_1+G_2}=I\frac{R_1}{R_1+R_2}$$

Inductors in series: According to KVL, the sum of voltages across the inductors is equal to the input voltage:

$$v=v_1+v_2+\cdots+v_n =L_1\frac{di}{dt}+\cdots+L_n\frac{di}{dt} =(L_1+L_2+\cdots+L_n)\frac{di}{dt} =L\frac{di}{dt}$$

i.e.,

$$L=L_1+L_2+\cdots+L_n$$

Inductors in parallel: According to KCL, the sum of currents through the inductors is equal to the input current:

$$i=i_1+i_2+\cdots+i_n =\frac{1}{L_1}\int v\; dt+\cdots+\frac{... ...cdots+\frac{1}{L_n}\right)\int v\; dt =\frac{1}{L}\int v\; dt$$

we get

$$\frac{1}{L}=\frac{1}{L_1}+\cdots+\frac{1}{L_n}$$

Capacitors in parallel: According to KCL, the sum of currents through the resistors is equal to the input current:

$$i=i_1+i_2+\cdots+i_n=C_1\frac{dv}{dt}+\cdots+C_n\frac{dv}{dt} =(C_1+C_2+\cdots+C_n)\frac{dv}{dt} =C\frac{dv}{dt}$$

i.e.,

$$C=C_1+C_2+\cdots+C_n$$

Capacitors in series: According to KVL, the sum of voltages across the capacitors is equal to the input voltage:

$$v=v_1+v_2+\cdots+v_n =\frac{1}{C_1}\int i\;dt+\cdots+\frac{1... ...+\cdots+\frac{1}{C_n}\right)\int i\;dt =\frac{1}{C}\int i\;dt$$

i.e.,

$$\frac{1}{C}=\frac{1}{C_1}+\cdots+\frac{1}{C_n}$$

	Resistor	Inductor	Capacitor
Governing Equation	$v=Ri$	$v=L\;di/dt$	$v=\int i\;dt/C$
Series connection	$R_s=R_1+R_2$	$L_s=L_1+L_2$	$C_s=1/C_1+1/C_2$
Parallel connection	$1/R_p=1/R_1+1/R_2$	$1/L_p=1/L_1+1/L_2$	$C_p=C_1+C_2$