Math:
(dot) webassign (dot) net / s e r c p 9 / 1 8 - p - 0 2 1 - a l t . g i f
Please find this question onto y a h o o s e a r c h e n g i n e b y c o p y & p a s t e.
Determine...
(a) the current in each resistor (Indicate the direction of the current flow through each resistor through the sign of your answer. Take upward current flow as positive.)
(b) the potential difference across the 200 Ohm resistor
_______________
In my physics class, the professor never gave us example of a circuit with multiple batteries. We have talked about simple circuits such as circuits making a single loop, but not something like this. I'm trying to learn from a very unhelpful textbook for hours now and I do not know what I am doing.
Any type of help you are willing to offer will be much appreciated!
5 answers
THE ABOVE IS THE LINK TO THE IMAGE WITHOUT ANY SPACES. AGAIN THANKS FOR YOUR HELP.
http://webassign.net/sercp9/18-p-021-alt.gif
E1 = 40V., R1 = 80 Ohms.
E2 = -360V., R2 = 20 Ohms.
E3 = 80V., R3 = 70 Ohms.
Ro = 200 Ohms.
Convert the 3 voltage sources to a single current source:
Is = E1/R1+E2/R2+E3/R3 = 40/80 - 360/20 + 80/70 = -16.9A., Downward.
Determine current through Ro:
1/Rs = 1/R1+1/R2+1/R3 = 1/80 + 1/20 + 1/70 = 0.07679, Rs = 13.0 Ohms.
Irs + Io = -16.9, (200/13)Io + Io = -16.9, 15.4Io + Io = -16.9,
Io = -1.03A.
Vo = Io*Ro = (-1.03) * 200 = -206 Volts.
Using the original circuit:
I1 = V1/R1 = (40+206)/80 = 3.075A.
I2 = V2/R2 = (-206+360)/20 = 7.7A.
I3 = V3/R3 = (80+206)/70 = 4.09A.
Note: The point where the 4 resistors are tied together is positive with respect to the opposite side.
E2 = -360V., R2 = 20 Ohms.
E3 = 80V., R3 = 70 Ohms.
Ro = 200 Ohms.
Convert the 3 voltage sources to a single current source:
Is = E1/R1+E2/R2+E3/R3 = 40/80 - 360/20 + 80/70 = -16.9A., Downward.
Determine current through Ro:
1/Rs = 1/R1+1/R2+1/R3 = 1/80 + 1/20 + 1/70 = 0.07679, Rs = 13.0 Ohms.
Irs + Io = -16.9, (200/13)Io + Io = -16.9, 15.4Io + Io = -16.9,
Io = -1.03A.
Vo = Io*Ro = (-1.03) * 200 = -206 Volts.
Using the original circuit:
I1 = V1/R1 = (40+206)/80 = 3.075A.
I2 = V2/R2 = (-206+360)/20 = 7.7A.
I3 = V3/R3 = (80+206)/70 = 4.09A.
Note: The point where the 4 resistors are tied together is positive with respect to the opposite side.
The above note does not apply to R2.