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Kirchhoff's Laws

Terminologies in electric circuits:

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

Kirchhoff's Laws

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

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

\begin{displaymath}\sum_k I_k=0 \end{displaymath}

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

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

\begin{displaymath}\sum_k V_k=0 \end{displaymath}

Kirchhoff.gif

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 $R_2$.

Kirchhoff1.gif

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

Apply KVL to the loop on the right to get:

\begin{displaymath}V_2-5+3=0,\;\;\;\;\;\; V_2=2V \end{displaymath}

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:

\begin{displaymath}2-I_1-I_2=2-I_1-1.5=0,\;\;\;\;\;I_1=0.5A \end{displaymath}

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

Apply KVL to the loop on the left to get:

\begin{displaymath}3\times 2 +5-V_0=0,\;\;\;\;\;\;V_0=11V \end{displaymath}

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

\begin{displaymath}V=V_1+V_2+\cdots+V_n=I(R_1+R_2+\cdots+R_n)=IR \end{displaymath}

i.e.,

\begin{displaymath}R=\frac{V}{I}=R_1+R_2+\cdots+R_n \end{displaymath}

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

\begin{displaymath}I=V/R_1+V/R_2+\cdots+V/R_n=V(1/R_1+1/R_2+\cdots+1/R_n)
=V/R \end{displaymath}

i.e.,

\begin{displaymath}R=\frac{V}{I}=\frac{1}{1/R_1+1/R_2+\cdots+1/R_n}=R_1//R_2//\cdots//R_n \end{displaymath}

In particular, when $n=2$,

\begin{displaymath}R=\frac{1}{1/R_1+1/R_2}=\frac{R_1\;R_2}{R_1+R_2} \end{displaymath}

Voltage Divider: From the voltage divider circuit, we see the output as the voltage across $R_2$ is

\begin{displaymath}V_2=R_2 I =R_2 \frac{V}{R_1+R_2}=V\frac{R_2}{R_1+R_2} \end{displaymath}

resistorseries.gif resistorparallel.gif

Current Divider: From the current divider circuit, we see the output as the current through $R_2$ is

\begin{displaymath}I_2=\frac{V}{R_2}=\frac{IR}{R_2}=I\frac{R_1R_2/(R_1+R_2)}{R_2}
=I\frac{R_1}{R_1+R_2} \end{displaymath}

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

\begin{displaymath}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} \end{displaymath}

i.e.,

\begin{displaymath}L=L_1+L_2+\cdots+L_n \end{displaymath}

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

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

we get

\begin{displaymath}\frac{1}{L}=\frac{1}{L_1}+\cdots+\frac{1}{L_n} \end{displaymath}

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

\begin{displaymath}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} \end{displaymath}

i.e.,

\begin{displaymath}C=C_1+C_2+\cdots+C_n \end{displaymath}

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

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

i.e.,

\begin{displaymath}\frac{1}{C}=\frac{1}{C_1}+\cdots+\frac{1}{C_n} \end{displaymath}

  Resistor Inductor Capacitor
Governing Equation $v=Ri$ $v=L\;di/dt$ $v=\int i\;dt/C $
Series connection $R_s=R_1+R_2$ $L_s=L_1+L_2$ $C_s=C_1C_2/(C_1+C_2)$
Parallel connection $R_p=R_1R_2/(R_1+R_2)$ $L_p=L_1L_2/(L_1+L_2)$ $C_p=C_1+C_2$


next up previous
Next: Energy Sources Up: Chapter 1: Basic Quantities Previous: Sinusoidal Variables
Ruye Wang 2008-02-08