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Next: Two-Port Networks Up: Chapter 2: Circuit Principles Previous: Solving Circuits with Kirchoff

Network Theorems

Example 1:

In the circuit below, $V_0=18V$, $R_1=R_2=3\Omega$, $R_3=6\Omega$, $R_4=1.5\Omega$. Find the value of current $I$ when $R_5$ is $1\Omega$, $2\Omega$, and $3\Omega$. Moreover, find the value for $R_5$ for the desired current $I=0.5 A$.

bridge_example.gif

Method 1, $\Delta-Y$ conversion

Find $I$ when $R_5=2\Omega$. First convert the $\Delta$ composed of $R_1$, $R_2$ and $R_5$ into a $Y$ composed of $R_a$, $R_b$ and $R_c$:

\begin{displaymath}R_c=\frac{R_1R_2}{R_1+R_2+R_5}=\frac{9}{8},\;\;\;\;
R_a=\frac{R_1R_5}{R_1+R_2+R_5}=\frac{3}{4}=R_b \end{displaymath}

Find overall resistance:

\begin{displaymath}R_{total}=R_c+(R_a+R_3) \vert\vert (R_b+R_4)=\frac{45}{16} \end{displaymath}

Find overall current:

\begin{displaymath}I_0=\frac{V_0}{R_{total}}=\frac{18}{45/16}=\frac{32}{5}\end{displaymath}

Find currents through $R_3$ and $R_4$ (current divider):

\begin{displaymath}I_a=I_0\;\frac{R_b+R_4}{(R_a+R_3)+(R_b+R_4)}=\frac{8}{5} \end{displaymath}


\begin{displaymath}I_b=I_0\;\frac{R_a+R_3}{(R_a+R_3)+(R_b+R_4)}=\frac{24}{5}\end{displaymath}

Find voltage at points $a$ and $b$ (assuming negative end of voltage source is ground):

\begin{displaymath}V_a=I_a \times R_3=\frac{8 }{5} \times 6=\frac{48}{5},\;\;\;\;\;
V_b=I_b \times R_4=\frac{24}{5} \times 1.5=\frac{36}{5} \end{displaymath}

Find current $I$ through $R_5=2\Omega$:

\begin{displaymath}I=\frac{V_a-V_b}{R_5}=\frac{12}{5}\times \frac{1}{2}=1.2\end{displaymath}

The same steps can be repeated for $R_5=1\Omega$ and $R_5=3\Omega$. But it is hard to find a value of $R_5$ given the require current $I=0.5$.

Method 2, Thevenin's theorem

Solve the problem using Thevenin's theorem by the following steps:

Here, we remove $R_5$ as the load of a network composed of all other resistors $R_1$, $R_2$, $R_3$, $R_4$ and the voltage source $V_0=18V$, then apply Thevenin's theorem to find the open-circuit voltage between the two terminals a and b:

\begin{displaymath}
V_T=V_{oc}=V_0\frac{R_3}{R_1+R_3}-V_0\frac{R_4}{R_2+R_4}
=18 \left(\frac{6}{9}-\frac{1.5}{4.5}\right)=6V
\end{displaymath}

and the internal resistance between a and b (with voltage source $V_0$ short-circuit):

\begin{displaymath}R_T=R_1//R_3+R_2//R_4=\frac{3\times 6}{3+6}+\frac{3\times 1.5}{3+1.5}=3\Omega \end{displaymath}

Now find current $I$ for different $R_5$ and when $I=0.5$, $R_5=V_T/I-R_T=6/0.5-3=9$

Example 2: The circuit below, often used in some control system, is composed of two voltages, two potentiometers, and a load resistor. Assume $V_1=72V$, $V_2=80V$, $R_1=1.5\Omega$, $R_2=3\Omega$, $R_3=1.5\Omega$, and $R_4=2.5\Omega$. Find the current $I_L$ through the load resistor $R_L=1.5\Omega$. We denote the current through $R_i$ by $I_i$, and the voltage at the left and right ends of $R_L$ by $V_a$ and $V_b$, respectively, with respect to the bottom wire treated as the ground.

potentiometers.gif

Method 1, Superposition theorem

Find $I'_L$ caused by voltage $V_1=72$, and then $I''_L$ caused by voltage $V_2=80$, then get $I_L=I'_L+I''_L$.

Method 2, Thevenin's theorem

Remove $R$, find open-circuit voltage $V_{ab}=V_a-V_b$ and equivalent resistance $R$, then find $I_L=V_{ab}/(R+R_L)$.


next up previous
Next: Two-Port Networks Up: Chapter 2: Circuit Principles Previous: Solving Circuits with Kirchoff
Ruye Wang 2014-02-09