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How would I calculate the voltage drop across the 60 ohm resistor without knowing the voltage? also, would the total circuit resistance be 7.74 ohms?

enter image description here

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closed as unclear what you're asking by Andy aka, Rev1.0, JIm Dearden, uint128_t, Daniel Grillo May 2 '16 at 12:23

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ "total circuit resistance" is meaningless unless two nodes are specified. \$\endgroup\$ – Andy aka Apr 30 '16 at 9:53
  • \$\begingroup\$ Also, which direction does the 1A flow and what value is "V"? Too many problems with this question - voting to close. \$\endgroup\$ – Andy aka Apr 30 '16 at 10:14
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    \$\begingroup\$ @Andyaka: "A" is not a current source, it's a meter. This gives the information needed to answer the question. Polarity doesn't matter. You can assume that "total resistance" is from the point of view of the voltage source. \$\endgroup\$ – Dave Tweed Apr 30 '16 at 10:21
  • \$\begingroup\$ It is not hard to figure out the currents: I will give you a starting hint. There is 1A flowing in a series branch with a 20 ohm and 10 ohm total. Given the standard rules, find the current in the 15 ohm resistor to the left; keep going left and you will find the current in the 60 ohm device. \$\endgroup\$ – Peter Smith Apr 30 '16 at 14:18
  • \$\begingroup\$ Does the direction of the current affect the approach to the question? \$\endgroup\$ – Genny Murphy Apr 30 '16 at 23:50
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I'm going to suggest the same thing I suggest every time questions involving lots of resistors arise. Circuits are topological, meaning that as long as you don't change which nodes are connected, you can draw the circuit however you like.

These questions are specifically formatted to be in a topology which is difficult to follow if you haven't yet built up the confidence to do so. Don't be afraid to redraw them into something more friendly. I tend to suggest you make everything vertical with sources increasing the distance of any node from the bottom of the diagram (they are adding voltage), and resistors decreasing the distance (they drop voltage). Furthermore, you can then unjumble the resistors to try and se which are in series and which are in parallel.

Here is the same diagram as the question redrawn. Hopefully now you can see where you went wrong on part (a).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Ok thank you. Would R4 and R7+R6 be in parallel? \$\endgroup\$ – Genny Murphy Apr 30 '16 at 23:46
  • \$\begingroup\$ @GennyMurphy yes. \$\endgroup\$ – Tom Carpenter May 1 '16 at 0:04
  • \$\begingroup\$ Not only a good reminder to think topologically sometimes, but adding the explicit links to CircuitLab and the saved simulation will be a helpful reminder of this capability for future SE questions. \$\endgroup\$ – uhoh May 2 '16 at 16:22
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    \$\begingroup\$ @uhoh the links under the schematic are inserted automatically. \$\endgroup\$ – Tom Carpenter May 2 '16 at 16:23
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We don't give direct answers to homework questions here — we expect you to show your work so far and explain exactly how you're stuck. But I'll give some hints:

How would I calculate the voltage drop across the 60 ohm resistor without knowing the voltage?

You don't need to know the voltage. The drop across the 60Ω resistor is going to be the same as the drop across the network to its right.

also, would the total circuit resistance be 7.74 ohms?

No. Isn't it obvious that it must be greater than 80Ω, because of the two series-connected 40Ω resistors?

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  • \$\begingroup\$ What I did was add the two 40 ohms together since they are in a series, and same with the 10 and 20 ohms. Then I ended up with 80 ohms, 60 ohms, 15 ohms and 30 ohms all parallel to eachother, and used the formula (1/Rt=1/R1 +1/R2 + 1/R3 + 1/R4) to come up with 7.74 ohms. Then I was left with 7.74 ohms and 5 ohms in series, which I added together to get 12.74 ohms all together. \$\endgroup\$ – Genny Murphy Apr 30 '16 at 13:13
  • \$\begingroup\$ @GennyMurphy You are right about the two sets of series resistors. However the rest is wrong. They aren't all in parallel with each other. \$\endgroup\$ – Tom Carpenter Apr 30 '16 at 15:09
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Simple..!! If 30 ohm resistor conducts 1 amp of current, the 15 ohm in parallel with it must able to conduct 2Amps with 30V across it. Then the 5 ohm must have 3A and 15V across it.

Thus across 60 ohm, voltage must be 45.

Total R is not difficult... Think of it.

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