Give the circuit as shown above. Detetmine E=? So that the 16V voltage source provides 32W power.
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2\$\begingroup\$ Two things. (1) There is a schematic editor available to you (mostly -- depends on cell phone, etc.) If you can use it, you should. There's little excuse providing a grainy picture like that. (2) You've opened the door to any and all techniques that might be applied. And given us nothing at all about what you already know about and can use. You should write something about how you think about these things. That will save us time and let us provide a more directly useful answer. What can you do with this, already? \$\endgroup\$– periblepsisCommented Jun 22 at 6:10
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\$\begingroup\$ Source transformations can often be used to solve problems. But here you will need to combine them with Wye-Delta transformations. Have you learned those yet? en.wikipedia.org/wiki/Y-%CE%94_transform \$\endgroup\$– Mattman944Commented Jun 22 at 7:37
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3\$\begingroup\$ This is a homework question and the OP needs to show more effort to solve it. SE isn't a homework answering service. \$\endgroup\$– Andy akaCommented Jun 22 at 10:45
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1 Answer
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Let me give you a start:
Your are told that \$V_1\$ provides \$32\:\text{W}\$ power. So this tells you the upper-right mesh current must be \$2\:\text{A}\$ and it must be positive in the indicated direction (counter-clockwise.) There's no way to avoid that fact.
From this, you must also know the current in \$R_1\$ is the same. And therefore, there must be a voltage drop across it. So you know the difference across \$R_1\$.
You are also given the upper-left mesh current because of the current source there.
I've added these two currents and their directions (when both are considered positive) to the above drawing.
Combining these two upper mesh currents should tell you exactly the current in \$R_2\$ and, therefore, also the exact voltage drop across \$R_2\$, as well.
Given these details, you should know the voltage drop across \$R_3\$. Just work yourself around the upper-right mesh loop using KVL. Knowing the voltage drop across \$R_3\$ tells you the total current in \$R_3\$, which is the sum of the currents in the two right-side meshes (upper and lower.)
So now you can work out the lower-right mesh current, which then tells you the voltage drop across \$R_4\$. With those in hand, you know the voltage drop across \$R_5\$, which tells you the total current in \$R_5\$. But that current is the sum of the two bottom mesh currents. And since you now know the mesh current in the lower-right mesh, this tells you the lower-left mesh current.
So you should have all the mesh currents worked out.
And from this, it's not difficult to find the voltage across \$E\$.
If it helps you, assign \$0\:\text{V}\$ to \$N_0\$. Then \$+16\:\text{V}\$ to \$N_1\$. Etc. Just work around the loops and place voltages, referenced to \$N_0\$, at each node.
