I really cannot solve this circuit.

[SeanHatch]

I've tried to solve isubx everyway I know. I know from the text that the answer is .4444 A, but I'd really like to know how to arrive at that answer. The latest thing I've tried is replacing the 2A source and the 10 ohm resistor (that are now in parallel) with a 20V source and the 10 ohm resistor in series. Then I can add the 10 ohm resistor to the 20 ohm to get a 30 ohm resistor just before the node. From there I'm not sure what to do. I've tried putting the current source back in, and I get a circuit with 2/3A for the current source, and then 3 branches of parallel 30 ohm resistors, which after current division gives me .22222. What am I doing wrong?

 

[instruite]

Use Kirchoff's current law and Currenbt division formula
KCL -
**broken link removed**
https://www.electronics2000.co.uk/data/itemsgl/Kirchoff.htm


Current Division formula -

 

[eblc1388]

Here is another way of doing it using Superposition.

Assume current in 15 Ohm resistor = 1A

Vb=30V and ib=1A also. So ix=2A.

Volt across 20 Ohm = 40V

Va=30+40=70V, ia=7A, total current I=9A

If current source=9A gives Ix=2A, then a 2A current source gives

Ix=2/9*2=0.4444A

 

[Nigel Goodwin]

Personally, I think the best thing to do is simplify it (just as you would with any algebra question), then once you've solved the complete circuit, work backwards from that to get the details.

For a start:

1) The two 15 ohm resistors are in series, so treat that at ONE 30 ohm resistor.

2) That then gives you two 30 ohm resistors in parallel, which equals 15 ohms.

3) That 15 ohms is in series with the 20 ohm resistor, giving a value of 35 ohms.

4) The 35 ohms is in parallel with the 10 ohm resistor, giving a total resistance for the circuit of 7.78 ohms (to two decimal places).

5) From ohms law this gives a voltage of 15.56V as the supply voltage (7.78x2).

6) Using ohms law again (15.56/35) gives roughly 0.444 amps current through the 20 ohm resistor.

7) This same current flows through the parallel 30 ohm resistors, so each 30 ohm resistor gets half the current = 0.222A.

8) As the 30 ohm 'resistor' we want is two 15 ohms in series, each will have the same current through them, so using ohms law again we get 3.3V across either 15 ohm (0.222x15).

I've worked this entirely for you (which as others will tell you ISN'T something I normally do!), but hopefully it will give you some insight in how to simplify such questions?.

 

[evandude]

another simple method is source transformations. you can transform the current source and parallel resistor into a voltage source with series resistor, then combine the two 15 ohm resistors and the 30 ohm, and then you end up with a series resistor loop with a voltage source, and can find isubx from that.

assuming you've covered source transformations in class... if not, you could probably learn it in 10 minutes by searching around online (it's not very complex) and impress your teacher

 

[SeanHatch]

Yeah we've learned source transformations, that's how I solved for the far resistor, but I must have made an error in solving for this current. Thanks to all who helped.

Sean