Solution for Example 5 (using Thevenin's theorem)

Remove $R$ as the load, model the rest of the circuit as a non-ideal Thevenin model in terms of $R_T$ and $V_T$. To find $R_T$, we short $V$ and find:

$$R_T=R_1\vert\vert(R_3+R_2\vert\vert R_5)=6\vert\vert(3+2\vert\vert 2)=6\vert\vert 4=2.4\,\Omega$$ (1)

To find $V_T$, we first find $V_5$ across $R_5$:

$$V_5=V\frac{R_5}{R_5+R_2\vert\vert(R_1+R_3)}=5 \frac{2}{2+2\vert\vert 9}=2.75\,V$$ (2)

$V_2$ across $R_2$:

$$V_5=V\frac{R_2\vert\vert(R_1+R_3)}{R_5+R_2\vert\vert(R_1+R_3)}=5 \frac{2}{2+2\vert\vert 9}=2.25\,V$$ (3)

$V_3$ across $R_3$:

$$V_3=V_2\frac{R_3}{R_1+R_3}=2.25\frac{3}{3+6}=0.75$$ (4)

Now we get $V_T=V_3+V_5=3.5$. To get $R$ so that the current through it is $I=0.5$, we solve the equation based on the Thevenin model

$$\frac{V_T}{R_T+R}=\frac{3.5}{2.4+R}=0.5$$ (5)

and get $R=4.6\,\Omega$.

Solution for Example 6

• $I_2$ goes through $R_4$ alone with a voltage drop $R_4I_2=10V$.
• $I_1$ splits between two parallel branches $R_2=15\Omega$ and $R_1+R_3=21\Omega$. The current through $R_1$ is
  

  $$I=I_1\frac{R_2}{R_1+R_2+R_3}=\frac{2\times 15}{15+21}=\frac{5}{6}A$$ (6)

  
  

• Voltage drop across $R_1$ is $R_1I=5V$.
• The open-circuit voltage across the output port is $V_T=V_{oc}=10+5+6-9=12V$.
• The total resistance between the two terminals is $R_T=R_{eq}=R_4+R_1\vert\vert(R_2+R_3)=10+6\vert\vert(15+15)=15\Omega$.