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Kirchhoff's Laws

Terminologies for electric circuits:

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.

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

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.

\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}

Voltage Divider:

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

\begin{displaymath}V_k=IR_k=\frac{V}{R_s}R_k=V\frac{R_k}{R_1+R_2+\cdots+R_n} \end{displaymath}

In particular if $n=2$, we have

\begin{displaymath}V_1=V\frac{R_1}{R_1+R_2},\;\;\;\;\;\;\;\;V_2=V\frac{R_2}{R_1+R_2} \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_s \end{displaymath}

i.e.,

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

resistorseries.gif

Resistors 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=\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} \end{displaymath}

i.e.,

\begin{displaymath}\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 \end{displaymath}

or

\begin{displaymath}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 \end{displaymath}

In particular, when $n=2$,

\begin{displaymath}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} \end{displaymath}

resistorparallel.gif

Current Divider:

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

\begin{displaymath}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} \end{displaymath}

In particular if $n=2$, we have

\begin{displaymath}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} \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{...
...cdots+\frac{1}{L_n}\right)\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...
...+\cdots+\frac{1}{C_n}\right)\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 $R_s=R_1+R_2$ $L_s=L_1+L_2$ $C_s=1/C_1+1/C_2$
Parallel connection $1/R_p=1/R_1+1/R_2$ $1/L_p=1/L_1+1/L_2$ $C_p=C_1+C_2$


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Next: Energy Sources Up: Chapter 1: Basic Quantities Previous: Sinusoidal Variables
Ruye Wang 2012-07-02