Am having a problem with this and I'm not sure what I'm doing incorrectly. Please see below the problem statement for what I've tried so far. Any guidance would be appreciated!
A 62.0 ohm resistor is connected in parallel with a 121.0 ohm resistor. This parallel group is connected in series with a 17.0 ohm resistor. The total combination is connected across a 15.0 V battery.
(a) Find the current in the 121.0 resistor.
(b) Find the power dissipated in the 121.0 resistor.
When I combined the parallel resistors I came up with 41.5 ohms which I then added to the 17 ohms that this resistors was in series with to get 58.5 ohms. However, when I used the equation I=V/R (I= 15.0V/58.5 ohms) the answer I got was not correct for the current. I'm not sure what I'm doing wrong here.
2007-02-02 09:19:31 · 6 answers · asked by larkinfan11 3 in Science & Mathematics ➔ Physics
Using 1/r1 + 1/r2 = 1/r(parallel)
You should get 40.994 ohms
+ 17
Gets 57.994 total ohms
The Voltage across the parallel resistors is 40.994/57.994 x 15
or 10.603 volts
I = V/R gives
I = 10.603 / 121
0.0876 amps
The current you calculated would be the total current across both the parallel resistors (added).
The power is of course 10.603 x 0.0876
or 0.929 watts.
2007-02-02 10:04:07 · answer #1 · answered by a simple man 6 · 0⤊ 0⤋
For the parallel combination I got 40.99 ohms, close to what you got but I'm wondering why it wasn't the same. Product over sum, right? 62*121/183.
But then when you calculated current that was way off? If I'm reading you right you expected 15.0V/58.5 ohms to be the current through the 121.0 resistor. That would be the current through the 17 ohm resistor but the current would then split, some throught the 121 and some through the 62.
Calculate the voltage drop across the 17 ohms, subtract that voltage from 15V. The result is the voltage across the parallel pair. Then use I=V/R again.
Then for b), P = I^2*R.
2007-02-02 17:36:34 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
(A) In a parallel circuit the voltage is constant throughout any amount of resistors therefore you have 15 volts at 62.0 ohm resistor and the 121 ohm resistor. So to find the current at the 121 ohm resister you take the 15 volts and devide by 121 ohms and you have your answer
(B) To find the power you can take the voltage and square it (15*15=225) and divide by 121 ohms and you have your answer.
2007-02-02 19:27:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The current is split between the two parallel resistors such that the voltage across the 62 ohm is the same as the voltage accross the 121 ohm:
121*I = 62*(15/85 - I)
121*I + 62*I = 62*15/85
183*I = 62*15/85
I = (62*15)/(121*85) = .0904 amps
The power dissipated is IR^2 = (.0904a)*(121^2ohm^2) = 1323.5 W
Bozo
2007-02-02 17:36:20 · answer #4 · answered by bozo 4 · 0⤊ 0⤋
First, get the total resistance. Mine's 57.99ohms.
Solve for It. I got 0.259A.
You can solve for the voltage for the 2 resistors in parallel connection by using the formula V=IR.
0.259*40.99=10.62
The voltages for each of the parallel resistors are equal to the total (only for the parallel part). So the 121 and 62 ohm resistors get 10.62 V each.
Through I=V/R, you can get I of 121 ohm resistor.
I=10.62/121=0.0878A
Then since P=IV
P=0.0878A*10.62V
P=0.932W
2007-02-02 21:33:04 · answer #5 · answered by crabmeat 2 · 0⤊ 0⤋
I get 40.995, but I am trusting the answer from a website.
You answer might be better or worse. Double check it.
When added to 17, then answer is 57.995.
2007-02-02 17:26:40 · answer #6 · answered by Anonymous · 0⤊ 0⤋