# Find the current and potential difference

by spice1510
Tags: current, difference, potential
 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
Emeritus
HW Helper
PF Gold
P: 7,235
 Quote by spice1510 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.
 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
Emeritus
HW Helper
PF Gold
P: 7,235

## Find the current and potential difference

 Quote by spice1510 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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