Need help with a circuit problem? (Picture attached)


by ObviousManiac
Tags: circuit, current, ohms law
ObviousManiac
ObviousManiac is offline
#1
Dec14-11, 09:39 PM
P: 37
**PICTURE ATTACHED AT BOTTOM**

1. The problem statement, all variables and given/known data
The current in the 13.8-Ω resistor in the attached picture is .795 A. Find the current in the other resistors in the circuit.

2. Relevant equations
V=RI

In a series circuit:
V of Circuit = V1 + V2 + V3 +...Vn
I of Circuit = I1 = I2 = I3 = ...In
R of Circuit = R1 + R2 + R3 +... Rn

In a parallel circuit:
V of Circuit = V1 = V2 = V3 = ... Vn
I of Circuit = I1 + I2 + I3 +... In
1/R of Circuit = 1/(R1) + 1/(R2) + 1/(R3) +... 1/(Rn)

3. The attempt at a solution

For the 13.8-Ω resistor, I found that Voltage = 10.971 by using V=IR
- Because V of Circuit = (V1 = V2 = V3 = ... Vn), the Voltage for the 17.2-Ω resistor is also 10.971
--Based on this, the Current for the 17.2-Ω resistor is .638:
V=IR
I=V/R
I=(10.971)/(17.2)
I=.638

After this, I proceeded to find the equivalent resistance for the entire circuit.

To find this, I separated the circuits into groups.

Group 1= the 15-Ω and 12.5-Ω resistors (they are in series).
Group 2= the 13.8-Ω and 17.2-Ω resistors (they are parallel).
Group 3= the 8.45-Ω and 4.11-Ω resistors (they are in series).

When solved for, the equivalent resistance for each group is as follows:

Group 1= 27.5-Ω
Group 2= 7.66-Ω
Group 3= 12.56-Ω

Note that Groups 2 and 3 are now parallel. We will combine them into Group 4.

Group 4 Resistance = 4.75-Ω

Now we only have Groups 1 and 4 left in series. When combined, we will get the net resistance for the entire Circuit. This works out to be 32.56-Ω



.....So that's everything I could find, which is decent, but doesn't come anywhere near answering the question. I know I must be missing something somewhere.

How do I figure out the individual currents for each resistor with the information I currently have?
Attached Thumbnails
Walker.21.44.jpg  
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gneill
gneill is offline
#2
Dec14-11, 10:14 PM
Mentor
P: 11,446
Quote Quote by ObviousManiac View Post
**PICTURE ATTACHED AT BOTTOM**

1. The problem statement, all variables and given/known data
The current in the 13.8-Ω resistor in the attached picture is .795 A. Find the current in the other resistors in the circuit.

2. Relevant equations
V=RI

In a series circuit:
V of Circuit = V1 + V2 + V3 +...Vn
I of Circuit = I1 = I2 = I3 = ...In
R of Circuit = R1 + R2 + R3 +... Rn

In a parallel circuit:
V of Circuit = V1 = V2 = V3 = ... Vn
I of Circuit = I1 + I2 + I3 +... In
1/R of Circuit = 1/(R1) + 1/(R2) + 1/(R3) +... 1/(Rn)

3. The attempt at a solution

For the 13.8-Ω resistor, I found that Voltage = 10.971 by using V=IR
- Because V of Circuit = (V1 = V2 = V3 = ... Vn), the Voltage for the 17.2-Ω resistor is also 10.971
--Based on this, the Current for the 17.2-Ω resistor is .638:
V=IR
I=V/R
I=(10.971)/(17.2)
I=.638

After this, I proceeded to find the equivalent resistance for the entire circuit.

<snip!>
You're doing so well using the voltage across the 13.8Ω resistor, why not continue? The same voltage must be across the series connected 8.45Ω + 4.11Ω resistors. So what's the current through them?

What then is the total current for the subcircuit?

Attached Thumbnails
Fig1.gif  
ObviousManiac
ObviousManiac is offline
#3
Dec14-11, 11:11 PM
P: 37
Ah, of course!

Okay, so based on that...

Current of the 8.45 Ω resistor should be (10.971)/(8.45) = 1.3 A

and the 4.11 Ω will be (10.971)/(4.11) = 2.67 A

So the total for the subcircuit will be

.795 A
+.638 A
+1.3 A
+2.67 A

= 5.403 A

....
Would that be correct?

gneill
gneill is offline
#4
Dec15-11, 07:22 AM
Mentor
P: 11,446

Need help with a circuit problem? (Picture attached)


Quote Quote by ObviousManiac View Post
Ah, of course!

Okay, so based on that...

Current of the 8.45 Ω resistor should be (10.971)/(8.45) = 1.3 A

and the 4.11 Ω will be (10.971)/(4.11) = 2.67 A

So the total for the subcircuit will be

.795 A
+.638 A
+1.3 A
+2.67 A

= 5.403 A

....
Would that be correct?
No, not quite. The 8.45 and 4.11 resistors are in series. How do you combine series resistances?
ObviousManiac
ObviousManiac is offline
#5
Dec15-11, 07:46 AM
P: 37
Quote Quote by gneill View Post
No, not quite. The 8.45 and 4.11 resistors are in series. How do you combine series resistances?
So would it be

(10.971)/(8.45+4.11) = .873 A

And then the subcircuit would work out as

.873 A
+ .795 A
+ .638 A

= 2.3 A
gneill
gneill is offline
#6
Dec15-11, 07:55 AM
Mentor
P: 11,446
Quote Quote by ObviousManiac View Post
So would it be

(10.971)/(8.45+4.11) = .873 A

And then the subcircuit would work out as

.873 A
+ .795 A
+ .638 A

= 2.3 A
Yessir. That looks good. You should now be in a position to state the current through every resistor in the circuit.
ObviousManiac
ObviousManiac is offline
#7
Dec15-11, 02:54 PM
P: 37
Alrighty, here are my final results. Everything seems to check out:

13 Ω = .795 A
17.2 Ω = .638 A
8.41 Ω = .87 A
4.11 Ω = .87 A
15 Ω = 2.3 A
12.56 Ω = 2.3 A
gneill
gneill is offline
#8
Dec15-11, 03:01 PM
Mentor
P: 11,446
Quote Quote by ObviousManiac View Post
Alrighty, here are my final results. Everything seems to check out:

13 Ω = .795 A
17.2 Ω = .638 A
8.41 Ω = .87 A
4.11 Ω = .87 A
15 Ω = 2.3 A
12.56 Ω = 2.3 A
Yup.
ObviousManiac
ObviousManiac is offline
#9
Dec15-11, 03:22 PM
P: 37
thanks for all the help.


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