Kirchhoff's Laws

Here are some terminologies for electric circuits:

• Node: a point where three or more current-carrying elements (branches) are connected;
• Branch: a path connecting two nodes along which there is one or more elements in series;
• Loop: a sequence of multiple paths that form a closed loop.

In a circuit diagram, both the direction of a current through, and the polarity of a voltage across an element can be arbitrarily labeled. The actual direction and polarity will be determined by the sign of the specific values obtained after the circuit is solved. For example, a current labeled in the left-to-right direction with a negative value is actually flowing in the right-to-left direction.

Kirchhoff's Laws

• Kirchhoff current Law (KCL)

  The algebraic sum of the currents into a node is zero:

  $$\sum_k I_k=0$$ (70)

  due to the principle of conservation of electric charge (electric charge can not be created or destroyed in the circuit).

  Here we can assume the directions of all currents through the elements are either into or out of the node.

• Kirchhoff voltage Law (KVL)

  The algebraic sum of all voltage drops around a loop is zero:

  $$\sum_k w_k=\sum_k V_k Q=0,\;\;\;\;\; \sum_k V_k=0$$ (71)

  due to the principle of conservation of energy (energy can not be created or destroyed in the circuit).

  Here we can assume the polarities of all voltages across the elements are from high (+) to low (-), while going around the loop in either clockwise or counter-clockwise direction.

Example 1:

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

Assume the current flows around the loop in clockwise direction, the KVL states: $-12+3+4+5=0$.

Example 2: Given the 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:

$$V_2-5+3=0,\;\;\;\;\;\; V_2=2V$$ (72)

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:

$$2-I_1-I_2=2-I_1-1.5=0,\;\;\;\;\;I_1=0.5A$$ (73)

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

Apply KVL to the loop on the left to get:

$$3\times 2 +5-V_0=0,\;\;\;\;\;\;V_0=11V$$ (74)

The series and parallel combinations of circuit components

• Resistors in series:

  

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

  $$V=V_1+\cdots+V_n = IR_1+\cdots+IR_n=I(R_1+\cdots+R_n)=IR_s$$

  $$= \frac{I}{G_1}+\cdots+\frac{I}{G_n} =I\left(\frac{1}{G_1}+\cdots+\frac{1}{G_n}\right)=\frac{I}{G_s}$$ (75)

  
  where

  $$R_s=\frac{V}{I}=R_1+\cdots+R_n$$ (76)

  and

  $$G_s=\frac{1}{R_s}=\frac{I}{V}=\frac{1}{1/G_1+\cdots+1/G_n}$$ (77)

• Voltage divider:

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

  $$V_k=IR_k=\frac{V}{R_s}R_k=V\frac{R_k}{R_1+R_2+\cdots+R_n}$$ (78)

  In particular if $n=2$, we have

  $$V_1=V\frac{R_1}{R_1+R_2},\;\;\;\;\;\;\;\;V_2=V\frac{R_2}{R_1+R_2}$$ (79)

• Resistors in parallel:

  

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

  $$I=I_1+\cdots+I_n = \frac{V}{R_1}+\cdots+\frac{V}{R_n} =V\left(\frac{1}{R_1}+\cdots+\frac{1}{R_n}\right)=\frac{V}{R_p}$$

  $$= VG_1+\cdots+VG_n=V(G_1+\cdots+G_n)=VG_p$$ (80)

  
  where

  $$R_p=\frac{V}{I}=\frac{1}{1/R_1+\cdots+1/R_n}=R_1\vert\vert R_2\vert\vert\cdots\vert\vert R_n$$ (81)

  and

  $$G_p=\frac{1}{R_p}=\frac{I}{V}=G_1+\cdots+G_n=\frac{1}{R_p} =\frac{1}{R_1}+\cdots+\frac{1}{R_n}$$ (82)

  In particular, when $n=2$,

  $$R_p=R_1\vert\vert R_2=\frac{1}{1/R_1+1/R_2}=\frac{R_1\;R_2}{R_1+R_2}$$ (83)

• Current divider:

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

  $$I_k=\frac{V}{R_k}=\frac{IR_p}{R_k}=I\frac{1/R_k}{1/R_1+1/R_2+\cdots+1/R_n} =I\frac{G_k}{G_1+G_2+\cdots+G_n}$$ (84)

  In particular if $n=2$, we have

  $$I_1=I\frac{G_1}{G_1+G_2}=I\frac{1/R_1}{1/R_1+1/R_2} =I\frac{R_2}{R_1+R_2}$$ (85)

  $$I_2=I\frac{G_2}{G_1+G_2}=I\frac{1/R_2}{1/R_1+1/R_2} =I\frac{R_1}{R_1+R_2}$$ (86)

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

  $$v=v_1+v_2+\cdots+v_n=L_1\frac{di}{dt}+\cdots+L_n\frac{di}{dt} =(L_1+L_2+\cdots+L_n)\frac{di}{dt}=L_s\frac{di}{dt}$$ (87)

  i.e.,

  $$L_s=L_1+L_2+\cdots+L_n$$ (88)

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

  $$i=i_1+i_2+\cdots+i_n =\frac{1}{L_1}\int v\; dt+\cdots+\frac{1}{L_... ...(\frac{1}{L_1}+\cdots+\frac{1}{L_n}\right)\int v\; dt =\frac{1}{L_p}\int v\; dt$$ (89)

  we get

  $$\frac{1}{L_p}=\frac{1}{L_1}+\cdots+\frac{1}{L_n}$$ (90)

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

  $$i=i_1+i_2+\cdots+i_n=C_1\frac{dv}{dt}+\cdots+C_n\frac{dv}{dt} =(C_1+C_2+\cdots+C_n)\frac{dv}{dt} =C_p\frac{dv}{dt}$$ (91)

  i.e.,

  $$C_p=C_1+C_2+\cdots+C_n$$ (92)

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

  $$v=v_1+v_2+\cdots+v_n =\frac{1}{C_1}\int i\;dt+\cdots+\frac{1}{C_n... ...ft(\frac{1}{C_1}+\cdots+\frac{1}{C_n}\right)\int i\;dt =\frac{1}{C_s}\int i\;dt$$ (93)

  i.e.,

  $$\frac{1}{C_s}=\frac{1}{C_1}+\cdots+\frac{1}{C_n}$$ (94)

	Resistor R	Inductor L	Capacitor C
Governing Equation	$v=Ri=i/G$, $i=v/R=Gv$	$v=L\;di/dt$, $i=\int v\; dt/L$	$v=\int i\;dt/C$, $i=C\;dv/dt$
Series connection	$R_s=R_1+R_2$, $1/G_s=1/G_1+1/G_2$	$L_s=L_1+L_2$	$1/C_s=1/C_1+1/C_2$
Parallel connection	$1/R_p=1/R_1+1/R_2$, $G_p=G_1+G_2$	$1/L_p=1/L_1+1/L_2$	$C_p=C_1+C_2$

Example 3 Consider the following six circuits as either current or voltage dividers.

• For each of the three parallel circuits, find $i_1$ and $i_2$ in terms of the given current $i$ and the resistances, capacitances, or inductances.
  • Resistor circuit:

    $$i_1=\frac{R_2}{R_1+R_2}i,\;\;\;i_2=\frac{R_1}{R_1+R_2}i$$ (95)

  • Capacitor circuit:

    $$i_1=\frac{C_1}{C_1+C_2}i,\;\;\;i_2=\frac{C_2}{C_1+C_2}i$$ (96)

  • Inductor circuit:

    $$i_1=\frac{L_2}{L_1+L_2}i,\;\;\;i_2=\frac{L_1}{L_1+L_2}i$$ (97)

• For each of the three series circuits, find $v_1$ and $v_2$ in terms of the given voltage $v$ and the resistances, capacitances, or inductances.
  • Resistor circuit:

    $$v_1=\frac{R_1}{R_1+R_2}v,\;\;\;v_2=\frac{R_2}{R_1+R_2}v$$ (98)

  • Capacitor circuit:

    $$v_1=\frac{C_2}{C_1+C_2}v,\;\;\;v_2=\frac{C_1}{C_1+C_2}v$$ (99)

  • Inductor circuit:

    $$v_1=\frac{L_1}{L_1+L_2}v,\;\;\;v_2=\frac{L_2}{L_1+L_2}v$$ (100)