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Find equivalent resistance, the current in R5, Voltage of R2 and power of R10.

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Actually, I don't know if my attempt is right. So, you would point out where there are mistakes, pleased.

R10 and R11 are in parallel (I called this R1011), also, R6 and R7 are in parallel (R67). Then R1011 and R12 are in parallel (R101112), then R9 and R101112 are in serie (I called it R9101112). Then R9101112 and R13 are in parallel (R910111213), then R1, R2, and R910111213 are in serie( R12910111213) And so on, Until now my circuit its this:

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Then are R5 and R3 in paralell? If this is right I can continue myself. Another question, how to get current in R5 and voltage of R2? Kirchhoff's Rules ?

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    \$\begingroup\$ Are the two ends of R5 and R3 connected to the same two nodes? If yes, then they're in parallel. \$\endgroup\$ – The Photon Nov 15 '19 at 19:39
  • \$\begingroup\$ Are you familiar with converting between Thevenin and Norton equivalents? This would simplify the analysis considerably. \$\endgroup\$ – Cristobol Polychronopolis Nov 15 '19 at 19:54
  • \$\begingroup\$ I would add some spacing character(s) between the resistor's numbers. How would you distinguish R112 from R112 (R11 in parallel with R2 resp. R1 in parallel with R12)? A commonly used seperator is || for parallel and + for series, e.g. (R1||R2)+R3... you need brackets as well... \$\endgroup\$ – Huisman Nov 15 '19 at 19:58
  • \$\begingroup\$ yea r5, r3 are parallel ! Then the ckt simplifies :-) \$\endgroup\$ – Meenie Leis Nov 16 '19 at 20:46
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There are many techniques that you could use to find the currents and voltages in the circuit, including the node-voltage method and the mesh-current method. In general, these methods will all rely on Kirchhoff's laws and Ohm's law.

Since you have only a resistor network and a single source you can combine resistors until you have a single resistor and the source...then find the current being provided by the source. That must be the same current as is flowing through R4, so now you know the voltage across R4. Work backwards through your simplifications of the resistor network, finding more detailed current and voltage information as you work your way back to the original circuit.

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