#### 3. Find the power absorbed by the 6V voltage source. The circuit labeled with "A" is unknown ![[Pasted image 20260324173748.png|center|450]] $R_{eq,2/3}=5\Omega\to I_{5\Omega}=\frac{V}{R}=\frac{5}{5}=1A$ $I_{10\Omega}=\frac{V}{R}=\frac{5}{10}=.5A$ $\text{Applying KCL @ 5V top node: }0.5+1+1.5=3A\to\text{ current thru 6V source is 3A upward}$ $\text{For 6V source, current is upward so it enters the negative terminal, leaves positive terminal }\therefore i=-3A$ $P=VI=-3(6)=-18\text{W}$ #### 5. In the circuit below, find the current $i_{x}$ ![[Pasted image 20260324174709.png|center|400]] $\text{Top node: }i_{1}=i_{2}+1~~~(1)$ $\text{Bottom node: }i_{5}=i_{4}+2~~~(2)$ $\text{Left-middle node: }i_{1}+i_{3}+i_{4}=0\to i_{1}=-(i_{3}+i_{4})~~~(3)$ $\text{Center node: }i_{x}=i_{2}+i_{3}+i_{5}$ $\text{Substituting (1)},(2),(3)\text{ into (4): }i_{x}=(-(i_{3}+i_{4})-1)+i_{3}+(i_{4}+2)=\dots=1A$ #### 6. Find $I_{AB}$ in the circuit below ![[Pasted image 20260324175213.png|center|450]] $\text{Node A: }\frac{V_{A}-50}{5}+\frac{V_{A}-100}{20}+\frac{V_{A}-V_{B}}{4}=0\to2V_{A}-V_{B}=60~~~(1)$ $\text{Node B: } \frac{V_{B}-V_{A}}{4}+(V_{B}-V_{C})-10=0~~~(2)$ $\text{Node C (far right): }(V_{C}-V_{B})+V_{C}+10-30=0\to 2V_{C}-V_{B}=20~~~(3)$ $\text{After substituting we get }V_{A}=52V,V_{B}=44V,V_{C}=32V$ $I_{AB}=\frac{V_{A}-V_{B}}{4}=\frac{52-44}{4}=2A$ #### 7. The circuit below contains a variable voltage source $V_{x}$, Find the value of $V_{x}$ such that the current $i_{x}$ is 2A ![[Pasted image 20260324180044.png|center|320]] $\text{Let node A be upper left, node B upper right, node C lower left, node D lower right}$ $V_{x}=2(1)=2V$ $V_{C}-V_{D}=0.5V$ $\text{Treat nodes A and B as supernode}$ $\frac{V_{A}-V_{C}}{1}=V_{A}-0.5, \frac{V_{B}-V_{D}}{1}=2\to V_{B}=2$ $\text{KCL: }(V_{A}-0.5)+2-1=0\to V_{A}=-0.5V$ $V_{x}=V_{B}-V_{A}=2-(-0.5)=2.5V$ #### 8. Find the Thevenin equivalent of the two-terminal network shown below ($V_{OC},R_{th}$) ![[Pasted image 20260324180503.png|center|300]] $i=0\therefore\text{ dependent source has value of }0V\text{ so current source forces 1A thru only available path}$ $\text{Output voltage is }V_{OC}=(1A)(1\Omega)=1V$ $\text{For }R_{th,}\text{ open circuit the independent source and use test source}$ $\text{Apply test current }i\text{ entering the port with direction shown in diagram}$ $\text{Dependent source drop is }V_{d}=2i\text{ so total port voltage is }V=2i+i=3i$ $R_{th}=\frac{V}{i}=3\Omega$