derfram
Favorite Answer
Resistors in series simply add
Resistors in parallel can be solved with 1/Rt = 1/R1 + 1/R2 + 1/R3....
or if only two resistors, then Rt = (R1 * R2)/(R1 + R2)
Once you combine all the resistors into one equivalent resistor across the battery you can use Ohm's law to determine the amount of current in the resistor.
Power = Voltage * Current.
Philomel
Req=2+2=4//4=2+10=14//12=168/26=6.46+4= 10.46Ω
P=E^2/R=25/10.46=
2.39W
Fireman
R6 & R7 are in series:
=>Resultant(Rr) = R6 + R7 = 2 + 2 = 4Ω
Now Rr & R5 are in parallel:
=>1/Rr = 1/4 + 1/4 => Rr = 2Ω
Now Rr and R4 are in series:
=>Rr1 = 2 + 10 = 12Ω
& R2 & R3 are also in series:
=>Rr2 = 4 + 8 = 12Ω
Now Rr1 & Rr2 are in parallel:
=>1/Rr = 1/12 + 1/12 =>Rr = 6Ω
Now Rr & R1 are in series:
=>R(net) = 4 + 6 = 10Ω
Thus by P = V^2/R
=>P(total) = (5)^2 /10 = 2.5 W
billrussell42
R6 and R7 and in series and add up to 4
that is in parallel with R5, and combine to 4/2 = 2
that is in series with R4, total 12
R2,3 in series add up to 12
that in parallel with the other 12 equals 6
and that in series with 4 equals 10 Ω
total current = 5/10 = 0.5 amps
total power = 5 x 0.5 = 2.5 watts
for R1, voltage drop is 2 volts, so power is 2x0.5 = 1 watt
0.5 amps into the network to the right of R1, which is 6 Ω, produces a voltage of 3 volts, which gives you a current into R2 and R3 of 3/12 = 1/4 amp
etc
etc
Old Science Guy
5,6,7 are in parallel
1/4 + 1/4 = 2/4 = 1/R so R = 2 ohm
that parallel combo is in series with 4
R = 2 + 10 = 12 ohm
all the above is in parallel with 2 and 3
1/12 + 1/12 = 2/12 = 1/R so R = 6 ohm
all the above is in series with 1
4 + 6 = 10 ohm total R
then
P = V I = 5v * 10 ohm = 50 watts
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