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!
Math:
(dot) webassign (dot) net / s e r c p 9 / 1 8 - p - 0 2 1 - a l t . g i f
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.
The above note does not apply to R2.
I'm sorry to hear that you're having trouble with your physics textbook. I'll do my best to explain how to approach this problem.
To solve this circuit problem, it's important to understand that there are two batteries in the circuit. The first step is to analyze the circuit and identify the components and their values. We have two resistors, one with an unknown value and the other with a resistance of 200 Ohms.
(a) To determine the current in each resistor, you need to apply Kirchhoff's laws. Kirchhoff's current law states that the sum of the currents entering a junction is equal to the sum of the currents leaving that junction. In this circuit, we have two junctions: one at the top and one at the bottom.
Let's assign variables to the currents in each resistor. Let's call the current flowing through the unknown resistor R1 and the current flowing through the 200 Ohm resistor as I2.
At the top junction, the total current entering the junction is the sum of the currents leaving the junction. Since there are two resistors in parallel, the potential difference across them is the same. Therefore, R1 has the same potential difference as the 200 Ohm resistor. Using Ohm's law (V = IR), we can say:
R1 = (V / R1) * R1
R1 = V
At the bottom junction, the total current entering the junction is the sum of the currents leaving the junction. The current leaving the junction is the same as the current flowing through the 200 Ohm resistor (I2). Therefore, we can write:
I2 = I1 + I3
Now, we need the additional information to calculate the values. If you can provide the voltage of the batteries or any other relevant data, we can proceed with the calculations.
(b) To determine the potential difference across the 200 Ohm resistor, we can use Ohm's law.
Potential Difference (V) = Current (I) * Resistance (R)
V = I2 * 200 Ohms
Unfortunately, without the necessary data, it's not possible to calculate the specific values in this circuit. I would recommend seeking additional resources, such as online tutorials or videos, to better understand circuit analysis with multiple batteries. Additionally, consider reaching out to your professor or classmates for clarification.