Find the voltage V in the circuit

[matinm90]

Homework Statement

we need to find V in this circuit. -see attachment

Homework Equations

how do I begin to simplify the circuit?

The Attempt at a Solution

 

[JamesGoh]

ok your given a 24 volt source in the middle.

by definition, any circuit component parallel to each other share the same voltage. this will mean that 24 volts will appear across the 12 ohm resistors in the middle.

from there you can try and determine the current and use Kirchoff current law to determine the loop currents. Hopefully you can then get the voltage you need

hope this helps

 

[matinm90]

funny u say that, my professor doesn't want us to use the kcl and kvl laws, basically no nodal or mesh analysis, which is why I am stuck, otherwise i can use those laws and figure it out in minutes.

 

[Zryn]

The diagram has been drawn so as to appear tricky, so redraw it with the source and each resistor laying vertical only for starters. You should only need the voltage divider equation (Ohms Law) to solve this.

 

[rude man]

1. Simplify the circuit: four of the resistors can immediately be reduced to two.
2. Compute the equivalent resistance of the two R's in series & the one in parallel with those two.
3. You now have 1 voltage source in series with two resistors.

 

[JamesGoh]

funny u say that, my professor doesn't want us to use the kcl and kvl laws, basically no nodal or mesh analysis, which is why I am stuck, otherwise i can use those laws and figure it out in minutes.

The 6 ohm on extreme left and 12 ohm resistor on extreme right are parallel to each other.

Finding the combined parallel resistance would be a good start

 

[NascentOxygen]

ok your given a 24 volt source in the middle.

by definition, any circuit component parallel to each other share the same voltage. this will mean that 24 volts will appear across the 12 ohm resistors in the middle.

You haven't noticed that 4 ohm resistor on the source's -ve terminal?

 

[JamesGoh]

You haven't noticed that 4 ohm resistor on the source's -ve terminal?

yes i have but the 6 ohm and 12 ohm on the extreme left and right of the circuit still share two common nodes on the circuit. By definition this makes them parallel

 

[Zryn]

by definition, any circuit component parallel to each other share the same voltage. this will mean that 24 volts will appear across the 12 ohm resistors in the middle.

NascentOxygen was commenting in regards to the bolded bit, which is incorrect, as the source is in series with the 4R resistor and this entire branch is then in parallel with the 12R resistors on either side of it, meaning that 24 volts will NOT appear across the 12R resistors as stated.

 

[gneill]

Sometimes it can be helpful to simply redraw the circuit, moving things about a bit (but not connections!) in order to bring out more familiar patterns. No analysis is required for this, just a bit of spatial imagination and manipulation.

For example, here's a version of the same circuit with things pushed about. Note that all the same component connections and nodes exist in this version; it is electrically identical to the original.

Parallel resistors are now obvious, so the obvious simplifications can be made there.

Those of you who know about Thevenin equivalents will recognize the opportunity to replace the part of the circuit left of red X's with its single source and resistor equivalent. Then you'll end up with a voltage source and three resistances in series. Easy-peasy.

Without Thevenin, KCL, or KVL, I suppose you can slog through the current calculations. Start by the net resistance seen by the voltage source and thus finding the total current that the source must drive through its 4 Ohm resistor, hence the voltage drop across it. This will tell you what the voltage is across the other two (parallel) branches. More current and voltage calculations... slog, slog, crank,... result.