Circuits with Series and Parallel Wiring

[Snape1830]

I attached the problem. I'm pretty confused. I remember the equations for parallel and series circuits, but I have no idea how to do this. I attempted, but failed. I know the two 10 ohm resistors are in parallel (so Rnet is 5), and the 20 ohm and the 6 ohm resistor are in series (So Rnet is 26). But then are those in parallel or series? And then i just don't know how to do the problem.

 

[vela]

The 10-ohm resistors are in parallel, but the 20-ohm and 6-ohm resistor aren't in series.

 

[Snape1830]

The 10-ohm resistors are in parallel, but the 20-ohm and 6-ohm resistor aren't in series.

Is the 10, 10 and 20 ohm in parallel and the 6 ohm in series with those three? Otherwise, I can't see how the 20 and the 6 aren't in series.

 

[vela]

The 20-ohm and 6-ohm are not in series because there's something else connected to the node where the two resistors are connected. If two elements are in series, all of the current in one element has to go through the other element. For example, the battery and the 6-ohm resistor are in series. You should be able to see that all of the current that flows through the battery has to also go through the 6-ohm resistor. That's not the case with the 6-ohm and 20-ohm resistor. Some of the current that goes through the 6-ohm resistor will branch off to go through the 10-ohm resistors.

 

[Snape1830]

The 20-ohm and 6-ohm are not in series because there's something else connected to the node where the two resistors are connected. If two elements are in series, all of the current in one element has to go through the other element. For example, the battery and the 6-ohm resistor are in series. You should be able to see that all of the current that flows through the battery has to also go through the 6-ohm resistor. That's not the case with the 6-ohm and 20-ohm resistor. Some of the current that goes through the 6-ohm resistor will branch off to go through the 10-ohm resistors.

Ok...so how do I do the problem?

 

[mrcheeses]

10 ohm resistors aren't in a series and are parallel but the other two are in a series. I find it easy to see if they are parallel or series by redrawing the diagram to one you are comfortable with. At every split in the wire with resistors, just branch it off.

 

[Snape1830]

10 ohm resistors aren't in a series and are parallel but the other two are in a series. I find it easy to see if they are parallel or series by redrawing the diagram to one you are comfortable with. At every split in the wire with resistors, just branch it off.

So 20 and 6 are in series? That's what I thought. I'm not really sure how to redraw it. Can you can just help me get the answer?

 

[vela]

Mrcheeses is wrong. The two resistors are not in series.

See http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Resist/Resist2.html

 

[ehild]

I redraw the circuit making the wires a bit shorter and moving the ammeter closer to the resistors, but the circuit is equivalent with the original one. Note that the ammeter can be replaced by a single piece of wire. Which resistors are connected in parallel?

ehild

 

[azizlwl]

Yes, 10,10 and 20 ohms are in parallel.

 

[Snape1830]

I redraw the circuit making the wires a bit shorter and moving the ammeter closer to the resistors, but the circuit is equivalent with the original one. Note that the ammeter can be replaced by a single piece of wire. Which resistors are connected in parallel?

ehild

10, 10 and the 20 ohm resistor are in parallel. So the Rnet is 4 for those? So then do I just use V=RI to solve for current?

 

[ehild]

Yes, the resultant of the parallel resistor is 4 ohm. And the resultant is connected to the 6 ohm resistor, and the whole is connected to the battery. You can use V=RI to get the total current.

What current does the ammeter read?

 

[Snape1830]

Yes, the resultant of the parallel resistor is 4 ohm. And the resultant is connected to the 6 ohm resistor, and the whole is connected to the battery. You can use V=RI to get the total current.

What current does the ammeter read?

Well, 4+6=10.
So...
20=10I
I=2 A

But 2 is wrong, so I don't know.

 

[ehild]

The ammeter does not read the current flowing through the battery. It reads the current that flows through itself. It is connected in series with what ?

ehild

 

[Snape1830]

The ammeter does not read the current flowing through the battery. It reads the current that flows through itself. It is connected in series with what ?

ehild

The three resistors in parallel? But to find current you need to know the voltage.

 

[ehild]

The three resistors in parallel? But to find current you need to know the voltage.

No it is not the three resistors in parallel, but?
And you can find the voltage. ehild

 

[Snape1830]

No it is not the three resistors in parallel, but?
And you can find the voltage.


ehild

Just the two 10s? How do I find the voltage if I only have the resistors?

 

[azizlwl]

Well, 4+6=10.
So...
20=10I
I=2 A

But 2 is wrong, so I don't know.

Now you got 2 Amps flowing through the equilvalent parallel resistors and through 6 Ohms resistor.

Now what is the voltage across each of the parallel resistors?
From this voltage drop, you calculate the current flow.

What is required here is to find the current flowing through the equivalent parallel resistors of two 10 Ohms resistors.

 

[Snape1830]

Now you got 2 Amps flowing through the equilvalent parallel resistors and through 6 Ohms resistor.

Now what is the voltage across each of the parallel resistors?
From this voltage drop, you calculate the current flow.

V-RI
V=10(2)
V=20 V
V=20(2)
V=40 V

So across the 10 ohm resistors, the voltage drop is 20 V
Across the 20 ohm resistor the voltage drop is 40 V?

That 40 can't be right, though?

 

[azizlwl]

Wrong.
With equivalent parallel resistors, you have only 2 resistors. 6 ohms and 4 ohms.
What is the voltage drop across 4 ohms resistor.
This voltage drop is identical to all the resistors in parallel.
Remember the voltage across all resistore in parallel are equal.

 

[Snape1830]

Wrong.
With equivalent parallel resistors, you have only 2 resistors. 6 ohms and 4 ohms.
What is the voltage drop across 4 ohms resistor.
This voltage drop is identical to all the resistors in parallel.
Remember the voltage across all resistore in parallel are equal.

Oh, right, I misread the question.
V=2(4)
V=8 Volts

 

[ehild]

Oh, right, I misread the question.
V=2(4)
V=8 Volts

So the voltage across the resultant of the parallel resistors is 8 V. You get the current through each of them if you divide that 8 V by the resistances.
What is the current that flows through the ammeter?
Remember Kirchhoff's nodal law.

ehild

 

[Snape1830]

So the voltage across the resultant of the parallel resistors is 8 V. You get the current through each of them if you divide that 8 V by the resistances.
What is the current that flows through the ammeter?
Remember Kirchhoff's nodal law.

ehild

Kirchoff's loop rule? The current going in equals the current coming out.
So do I find the current through of the parallel resistors and then add them up...because that comes out to 2 and 2 is wrong.

 

[ehild]

Kirchoff's loop rule? The current going in equals the current coming out.
So do I find the current through of the parallel resistors and then add them up...because that comes out to 2 and 2 is wrong.

I wrote Kirchhoff's node rule.

The ammeter measures the current that flows through it. Does I₂₀ ,represented by the green arrow flow through the ammeter?

ehild

 

[Snape1830]

Wrote Kirchhoff's node rule.

The ammeter measures the current that flows through it. Does I₂₀ ,represented by the green arrow flow through the ammeter?

ehild

No? So it would be 1.6 A?

And thanks for taking the time to edit the picture!

 

[ehild]

No? So it would be 1.6 A?

Correct .

ehild

 

[Snape1830]

Correct .

ehild

Ok! So now part b how much power dissipated through a 10 ohm resistor. The power equation.
P=IV
or
P=I²(r)
or
P= v²/R

I tried a few answers but none of them work.

 

[ehild]

What is the voltage across the 10 ohm resistor?

ehild

 

[Snape1830]

What is the voltage across the 10 ohm resistor?

ehild

8? It seems wrong for some reason

 

[NascentOxygen]

Ok! So now part b how much power dissipated through a 10 ohm resistor. The power equation.
P=IV
or
P=I²(r)
or
P= v²/R

I tried a few answers but none of them work.

They don't? http://img690.imageshack.us/img690/3484/hidden17.gif

How much current is flowing in just one of the 10Ω resistors?

 

[Snape1830]

They don't? http://img690.imageshack.us/img690/3484/hidden17.gif

How much current is flowing in just one of the 10Ω resistors?

.8 amps?

 

[NascentOxygen]

.8 amps?

Then I²·R =
and what makes you think this isn't right?

 

[Snape1830]

Then I²·R =
and what makes you think this isn't right?

It would be 6.4, and it's wrong because I plugged in the answer and it said it was wrong.

 

[NascentOxygen]

It would be 6.4, and it's wrong because I plugged in the answer and it said it was wrong.

6.4 watts is correct.

 

[Snape1830]

6.4 watts is correct.

Oh, right yeah, I just didn't put the decimal point in. Silly me.

Thanks!

 

[Rajatmo]

questions like these can be solved in your head!
first, for the current, the net resistance of the topmost two resistors is 5 ohm. now, the reciprocal of 5 is .2 and the reciprocal of 20 is .05 as you should be able to do immediately. they add up to be .25 - which is the reciprocal of the net resistance of the equivalent resistor containing the 20 ohm resistor and also the reciprocal of 4. so, to date, the net resistance is 4 ohms and this adds to the 6 ohms resistor in series. now, clearly, the net current through the battery is 20/(6+4) A = 2 A. this 2 A current divides in two sections in the ratio 4:1 (the current varies inversely as the resistance with p.d. constant). so, the current going through the ammeter is 2\times4/5 = 16/10 = 1.6 A.
for the power, the current 1.6 A just divides in two equal parts, so it's .8 A through each 10 A resistor. so, the power dissipated is 10\times.8² = 6.4 W.

so, don't use the loop and the junction rules unless you really need them. they'll mess up everything. try to solve this kinds of problems intuitively or using the method I've described - they are really easy.