Problem with current and parallel resistors.

[KingJaymz]

Problem with current and parallel resistors. Please help!

Ok, I've got this homework problem that I'm somewhat stuck with. The problem relates to the picture below:

I've got to figure out the total resistivity (done that), total current (done that), the current through the 21 ohm resistor and the voltage drop at the 21 ohm resistor. So I figured I'd find the current first, but I've got a problem. Once the circuit comes back together from the 4 and 17 ohm resistors, how do I calculate the current? I can't find any examples in my book of the resistance of a resistor that comes after a parallel circuit that has been closed again. Please help!

 

[formulajoe]

ya got to use current divider.
its like this

I = [(overall current) X other resistor]/R1 + R2

 

[KingJaymz]

But the resistors that are a problem (4 and 17 ohms) aren't in series, so how can I add them?

 

[formulajoe]

what you are doing is determing the amount of the current that will flow through each resister correct?
Current divider is a proportional equation. most of the current is going to take the path of least resistance, so the 4 ohm resister is going to have a greater current.

thats just the equation.

 

[KingJaymz]

Right, but the cables from the 4 and 17 ohm resistors meet again before heading on to the 21 ohm resistor. So I'll need the current at the joint (at least for the voltage drop at the 21 ohm resistor), right?

 

[formulajoe]

the current going into the parallel network will be the same when it comes out. looks like your biggest problem is figuring out how much branches off into the the 12 ohm resister.

 

[KingJaymz]

Thank you. That's what I needed to know. So can I treat the 4 and 17 ohm resistors as one resistor if I use the combined current value? Not sure if I worded that correctly, hopefully I did.

 

[chroot]

Yes. For the sake of finding the current through the 21 ohm resistor, the 4 and 17 ohm resistors can be considered combined as one.

- Warren

 

[KingJaymz]

Thank you. I thought that might be the case in the beginning, but I wasn't entirely sure. Thank you again.

 

[KingJaymz]

Another question came up in the same problem. I guess I don't understand exactly how to calculate the current through the 10 ohm resistor. I get 12 amps doing it the way I thought would be right, but that's a whole lot larger than the total current (4.61). Are the current values not additive? I'm guessing that my problem here lies in the fact that I'm confused as to how to accurately calculate the voltage drop. I can't use the total current, and I'm not sure of a voltage drop equation. Thank you for your help. You all are very helpful.

 

[formulajoe]

since the 10 ohm resister is in series with the voltage source, it will have the same current going through it. to find the voltage drop, use ohms law. V = IR.
i came up with the same overall curren that you got. so, the current through the 10 ohm resister is 4.61 Amps. the voltage is going to be (10 ohms)*(4.61 Amps) = 46.1 Volts.

 

[KingJaymz]

Ah, I see. Thank you so much for your help.

 

[chroot]

Step 1: Find total resistance. 4 || 17 = 68/21. 12 || (68/21 + 21) = 6801/761. 10 + 8 + 6801/761 = 19806/761, or about 26 ohms.

120 V into 19806/761 ohms is 15220/3301 A, or about 4.61 A. Check.

Current through 10 ohm resistor is 4.61 A. Voltage drop is (4.61)(10) = 46.1 V.

Current through 8 ohm resistor is 4.61 A. Voltage drop is (4.61)(8) = 36.88 V.

Voltage across the entire bunch of 4 resistors in the middle is (120 - 46.1 - 36.88 = 37 V.

Current through 12 ohm resistor is 37 V / 12 ohms = 3.08 A.

Current through 4, 17, and 21 ohm resistors combined is (4.61 - 3.08 = 1.53 A).

Current through 21 ohm resistor is 1.53 A. Voltage across 21 ohm resistor is 1.53 * 21 = 32.07 V.

- Warren

 

[KingJaymz]

Thank you! It's great to be able to verify that I got the right answers.