Circuit problem, prob really easy, please check

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The discussion focuses on finding missing values in a circuit, with an initial calculation of total resistance at 1.4 ohms. It clarifies that the resistances are not all in parallel and suggests using Kirchhoff's laws alongside Ohm's law to derive simultaneous equations for the currents. A mention of Thevenin's Theorem offers a quicker method to simplify the circuit analysis, although the original poster has not yet covered this topic. The explanation emphasizes that drawing additional voltage sources and replacing components with their Thevenin equivalents can lead to a simpler series circuit. Overall, the conversation provides insights into circuit analysis techniques and problem-solving strategies.
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1. Find the missing values for the given circuit.



2. the given circuit is attached..
View attachment phys.bmp



3. soo, so far, i used ohms laws to get the total resistance, Rt = 1.4 ohms. can someone please tell me if this is right?
 
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are all of them in parallel here ?
 
The resistance of the whole circuit is indeed 1.4 ohm. The resistances are NOT all in parallel.
In general The resistance of every network of resistances can be computed by using kirchhofss laws and ohms law to get simultaneous equations for all the currents in the circuit.
You can use kirchhofs law for every 3 or more wire connection
I_1 = I_3 + i_4 for example
and ohms law and the fact that the sum of all voltages differences in a closed loop
is 0
V = R_1*I_1 + R_4*I_4
V = R_1*I_1 + R_3*I_3+R_5*I_5
etc.
(use always the same direction for I_3)
If you use enough of those you can get 5 simultaneous equations for I_1 ... I_5

There's a faster way, if you've covered Thevenins Theorem.

http://en.wikipedia.org/wiki/Thévenin's_theorem

draw another 14V voltage source to the right of R2 and R5 and parallel to the first voltage source. This won't change any currents through the resistances.
Then cut the wire between the tops of R1 and R2. Both sides of the cut will be at the same voltage before and after, so that won't change any currents either.
Then replace the left voltage source, R1 and R4 with their thevenin equivalent and do the
same with the voltage source on the right, R2 and R5.
What remains is a simple series circuit.
 
thank youu.
we haven't covered thevenin yet but wat u said made sense =)
 
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