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Question

I am stuck on this problem and would be really greatful with any help on it.

Find the current in the 12 Ω resistor shown below (where: R1 = 2.2 Ω, R2 = 8.6 Ω, V = 24 V).

http://www.webassign.net/sercp/p18-13alt.gif

Answer 1

I won't do the problem for you but here is where to start:

Use the rules for calculating effective Resistance of resistors in series and parallel to collapse the circuit into a more simpler one. Begin with combining R2, R2, and R1. Once you have the circuit as simplified as you can get it, use ohms law to calculate the current.

Answer 2

Okay, first thing you need to do is print out that diagram, then get a pencil, then start writing down little numbers that show what the voltage is at various key points in the circuit.

Hint: Write "0 V" on the left side of the battery icon; and "24 V" on the right side of the battery icon.

Another hint: As you move along the circuit, as long as you're just moving along PLAIN WIRE (the grey lines), the voltage doesn't change. The voltage only changes when you cross from one side of a resistor to the other, or from one side of the battery to the other.

So, given that 2nd hint, you should immediately be able to write down "0 V" in at least one more place, and "24 V" in several more places. Do that now.

Now you need to start figuring out the current that's flowing across various sections of the circuit. This is easy when there's only one resistor, because then the formula is:

I = ΔV/R

where ΔV is the voltage change from one side of the resistor to the other; and "R" is the resistor's resistance.

But you're problem will take more work, because you have multiple resistors. When you have a big lump of resistors, and you know the voltage difference (ΔV) from one side of the lump to the other side of the lump, then the current flowing through the whole lump is:

I = ΔV/(effective resistance)

where (effective resistance) is a combination of the resistances of all the resistors in the lump.

Now, you DO know how to find the effective resistance of two resistors in series, and the effective resistance of two resistors in parallel, right? If so, then you should (by doing several calculations) be able to figure out the effective resistance of the big lump in the diagram (you'll probably have to break it into several smaller lumps first).

If you DON'T know how to do those series- and parallel-calculations, you are just not ready for this question yet. Crack open your book and find those formulas.

Anyway, once you know some of the currents, you should start writing those down too. Every time you figure out the current that's going through some resistor, write it down next to the resistor (write an "A" for "Amperes" next to it; so you'll know it means current; not voltage or resistance).

Now, if you know the current that's going through some resistor, and you know the voltage on one side of the resistor, you can figure out the voltage that's on the other side. You can use the exact same formula I gave above, just rearranged:

ΔV = (change in voltage) = IR

(Hint: Voltage always _drops_ in the direction of the current flow. So also draw arrows to show which way the current is flowing through the resistors).

Anyway, that's basically all you need to know. Now your job is to, one by one, fill in the voltaga values on the left and right side of each resistor, and fill in the current values traveling through each resistor. One of those will be the 12&Ohm; resistor, which will answer your problem.

Answer 3

Combine all the resistors.

click on the link to see the diagram:
http://i120.photobucket.com/albums/o188/stmc247/untitled.jpg

To find the current at 12Ω. You need to find the voltage drop at 12Ω, then use I = V/R to find the current. To do this, you need to work backward.

The voltage drop at 5.03Ω is 24V, which is the total voltage drop of the whole circuit.

5.03Ω is the combination of 2.2Ω and 2.83Ω, thus, the voltage drop at these two is 10.494V and 13.5V.


2.83Ω is the combination of 6.5Ω and 5Ω. The voltage drop at these two resistors is 13.5V.

Now, you need to find the current through 5Ω.
I = V/R = 13.5 / 5 = 2.7A

5Ω is the combination of 2Ω and 3Ω. Thus, the current through these two resistors is 2.7A

Now you need to find the voltage drop and the 3Ω.

V = RI = 3*2.7 = 8.1V

3Ω is the combination of 4Ω and 12Ω. Because 4Ω and 12Ω are in parrellel postion, the voltage drop throught these two resistors is also 8.1V. Therefore, the voltage drop at 12Ω is 8.1V.

I = V/R

I = 8.1V / 12Ω

I = .675A

Answer 4

In a series circuit the current would be 2.2 amperes.