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Circuits with Series and Parallel Wiring

 
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Jun9-12, 07:35 PM   #18
 

Circuits with Series and Parallel Wiring


Quote by Snape1830 View Post
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.
 
Jun9-12, 07:43 PM   #19
 
Quote by azizlwl View Post
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?
 
Jun9-12, 07:50 PM   #20
 
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.
 
Jun9-12, 08:38 PM   #21
 
Quote by azizlwl View Post
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
 
Jun9-12, 10:43 PM   #22

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Quote by Snape1830 View Post
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
Attached Thumbnails
circuit3b.JPG  
 
Jun10-12, 08:18 AM   #23
 
Quote by ehild View Post
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.
 
Jun10-12, 08:25 AM   #24

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Quote by Snape1830 View Post
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
 
Jun10-12, 08:30 AM   #25
 
Quote by ehild View Post
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!
 
Jun10-12, 08:41 AM   #26

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Quote by Snape1830 View Post
No? So it would be 1.6 A?

Correct .

ehild
 
Jun10-12, 08:52 AM   #27
 
Quote by ehild View Post
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.
 
Jun10-12, 09:16 AM   #28

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What is the voltage across the 10 ohm resistor?

ehild
 
Jun10-12, 09:18 AM   #29
 
Quote by ehild View Post
What is the voltage across the 10 ohm resistor?

ehild
8? It seems wrong for some reason
 
Jun10-12, 09:18 AM   #30
 
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Quote by Snape1830 View Post
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?

How much current is flowing in just one of the 10Ω resistors?
 
Jun10-12, 09:22 AM   #31
 
Quote by NascentOxygen View Post
They don't?

How much current is flowing in just one of the 10Ω resistors?
.8 amps?
 
Jun10-12, 09:26 AM   #32
 
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Quote by Snape1830 View Post
.8 amps?
Then I2·R =
and what makes you think this isn't right?
 
Jun10-12, 09:27 AM   #33
 
Quote by NascentOxygen View Post
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.
 
Jun10-12, 10:29 AM   #34
 
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Quote by Snape1830 View Post
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.
 
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