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Find the current and potential difference

  • Thread starter spice1510
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1. Homework Statement
For the circuit shown in the figure, find the current through and the potential difference across each resistor.
http://session.masteringphysics.com/problemAsset/1169464/1/jfk.Figure.23.P60.jpg


2. Homework 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. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,188
930
1. Homework Statement
For the circuit shown in the figure, find the current through and the potential difference across each resistor.
http://session.masteringphysics.com/problemAsset/1169464/1/jfk.Figure.23.P60.jpg


2. Homework 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. Homework Statement

2. Homework Equations

3. The Attempt at a Solution
See the comment in red.
 
2
0
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
 
Last edited:

SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,188
930
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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