Find the current and potential difference


by spice1510
Tags: current, difference, potential
spice1510
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#1
Mar11-11, 08:23 PM
P: 2
1. The problem statement, all variables and given/known data
For the circuit shown in the figure, find the current through and the potential difference across each resistor.
http://session.masteringphysics.com/...ure.23.P60.jpg


2. Relevant equations
V=I(amps)/R(ohms) and Req=R1+R2+R3.....


3. The attempt at a solution
Rearranging the circuit:
R(8ohm) and R(24ohm) are parallel so (1/8+1/24)^-1=6ohm
that created a series of R(6ohm)+R(6ohm) so R6+R6=R(12ohm)
now I have the first circuit R(4ohm) in parallel with R(12ohm) and R(24ohm)
(1/4+1/12+1/24)^-1=2.67ohm for equivalent resistance for the entire circuit

24V=I*2.67ohm
I=9Amps

across the 4ohm resistor the current is? I=9A because the current is the same or is it 9A divided across the 3 parallel resistors?

I'm either getting it wrong because I put the 4ohm by itself (before the junction correct?) or if it's in series with the 2 6ohm resistors then that's my error but I've definitely tried it 4 different ways and am getting something wrong.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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SammyS
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#2
Mar11-11, 09:02 PM
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Quote Quote by spice1510 View Post
1. The problem statement, all variables and given/known data
For the circuit shown in the figure, find the current through and the potential difference across each resistor.
http://session.masteringphysics.com/...ure.23.P60.jpg


2. Relevant equations
V=I(amps)/R(ohms) and Req=R1+R2+R3.....


3. The attempt at a solution
Rearranging the circuit:
R(8ohm) and R(24ohm) are parallel so (1/8+1/24)^-1=6ohm
that created a series of R(6ohm)+R(6ohm) so R6+R6=R(12ohm)
now I have the first circuit R(4ohm) in parallel with R(12ohm) and R(24ohm)

No, the 12Ω and 24Ω are in parallel. The 4V is in series with the (12V & 24V) combination.

(1/4+1/12+1/24)^-1=2.67ohm for equivalent resistance for the entire circuit

24V=I*2.67ohm
I=9Amps

across the 4ohm resistor the current is? I=9A because the current is the same or is it 9A divided across the 3 parallel resistors?

I'm either getting it wrong because I put the 4ohm by itself (before the junction correct?) or if it's in series with the 2 6ohm resistors then that's my error but I've definitely tried it 4 different ways and am getting something wrong.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
See the comment in red.
spice1510
spice1510 is offline
#3
Mar11-11, 09:16 PM
P: 2
so 4ohm in series with (1/12+1/24)^-1 which equals 8ohm

4ohm+8ohm (series)=12ohm for the entire circuit

24V=12ohm*I
I=2amps

SammyS
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#4
Mar12-11, 11:18 PM
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P: 7,379

Find the current and potential difference


Quote Quote by spice1510 View Post
so 4ohm in series with (1/12+1/24)^-1 which equals 8ohm

4ohm+8ohm (series)=12ohm for the entire circuit

24V=12ohm*I
I=2amps
Yes.

Now, work your way back through all the resistors.


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