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Definitely YES!
O.k, o.k, I'm aware that this is not the answer you expected. And it isn't the least helpful.
Here's the (hopefully) helpful part:
- start by teh series connection of R5+R6 by a single resistor (let's call it R7). You should know how to calculate this series connection.
- next replace the parallel circuit of R4 and the newly created R7 by another single resistor (let's call it R8). Yiu should know how to calculate this parallel connection, too.
- Go on in the same way until you are left with R1 and R10. You now have a voltage divider that can be calculated easily.
Now work you way backwards using e.g. Thevenin's theorem.
Another method would set up all the node equations using kirchhoffs laws and solve the resulting equation system.
Hint: V(R6)=9.511V (if I'm not mistaken)
Here's the (hopefully) helpful part:
- start by teh series connection of R5+R6 by a single resistor (let's call it R7). You should know how to calculate this series connection.
- next replace the parallel circuit of R4 and the newly created R7 by another single resistor (let's call it R8). Yiu should know how to calculate this parallel connection, too.
- Go on in the same way until you are left with R1 and R10. You now have a voltage divider that can be calculated easily.
Now work you way backwards using e.g. Thevenin's theorem.
Another method would set up all the node equations using kirchhoffs laws and solve the resulting equation system.
Hint: V(R6)=9.511V (if I'm not mistaken)
