A battery with voltage drop of 120V is connected in a ciruit containing both series and parallel. Moving in a clockwise direction. R1= 3 ohms. Suddenly there is a connection of point A to point B. In between these two points is a parallel circuit with R2= 10 ohms and R3=2ohms on the top(these two are in series with each other). On the bottom of the rectangle there is a parallel circuit consisting of R4= 20 ohms and R5= 5 ohms. After point B is one more resistor R6= 6 ohms. (So the image is one large rectange and on the top length of the rectange are all of the resistor. With R1 on the far left and R6 on the far right. Between R1 and R2 is anohter rectange, where on the top length are R2 and R3(in series with each other). Then there is one more small square on the bottom length of the second rectange where R4 and R5 are in parallel with each other. )Find the voltage drop and current across each resistor.
I know that the voltage drop between points A and B is 30V because I know that V-total - V1 - V-6 = the voltage drop between point A and B.
What are the current and voltage drop of the other remaining resistors(R2, R3, R4, and R5)?V=IR
I have been trying for hours to solve this problem but I just can't get the right answer. I can only figure out the information for R1(10A and 30V) and R6(10A and 60V). Can you help me with the rest?
can you post a drawing on the web? If so, please repost. You will have to put spaces in your web address, as in
w w w. P i c t u r e H e r e. c o m
Sure, I can help you with the remaining resistors. Let's solve the problem step by step.
1. Voltage drop across R1:
You mentioned that the total voltage drop between points A and B is 30V. Therefore, the voltage drop across R1 (V1) can be calculated by subtracting the voltage drop across R6 (V6) from the total voltage drop. V1 = V_total - V6 = 30V - 60V = -30V (negative sign indicates the direction is opposite to the clockwise direction).
2. Current across R1:
We can use Ohm's Law to find the current (I1) flowing through R1. I1 = V1 / R1 = -30V / 3Ω = -10A (Amps). Again, the negative sign indicates the direction is opposite to the clockwise direction.
3. Current across R2 and R3:
Since R2 and R3 are connected in series, the current flowing through both resistors will be the same. Let's call it I2. To find I2, we can use the formula I2 = V2 / (R2 + R3), where V2 is the voltage drop across the combination of R2 and R3. We need to find V2 first.
To find V2, we can use the parallel voltage divider rule. Consider R2 and R3 as a combination, and let's call it R23. The formula for finding the voltage drop across R23 (V23) is V23 = (R3 / (R2 + R3)) * V_total. Plugging in the values, we get V23 = (2Ω / (10Ω + 2Ω)) * 30V = (2/12) * 30V = 5V.
Since R2 and R3 are in series, the voltage drop across each of them will be the same as V23, which is 5V.
Now, we can calculate I2 using the formula I2 = V2 / (R2 + R3). I2 = 5V / (10Ω + 2Ω) = 5V / 12Ω = 0.417A (approximately).
4. Voltage drop across R4 and R5:
Since R4 and R5 are connected in parallel, the voltage drop across both resistors will be the same. Let's call it V4,5. To find V4,5, we can use the parallel voltage divider rule. The formula is V4,5 = (R4 / (R4 + R5)) * V_total. Plugging in the values, we get V4,5 = (20Ω / (20Ω + 5Ω)) * 30V = (20/25) * 30V = 24V.
This means the voltage drop across both R4 and R5 is 24V.
5. Current across R4 and R5:
Since the voltage drop across R4 and R5 is the same, the current flowing through both resistors will also be the same. Let's call it I4,5. To find I4,5, we can use Ohm's Law. I4,5 = V4,5 / R4 = 24V / 20Ω = 1.2A.
6. Current across R6:
The current across R6 (I6) can be calculated using Ohm's Law. I6 = V6 / R6 = 60V / 6Ω = 10A.
So, to summarize:
- Voltage drop and current across R1: V1 = -30V, I1 = -10A
- Voltage drop and current across R2 and R3: V2 = 5V, I2 = 0.417A
- Voltage drop and current across R4 and R5: V4,5 = 24V, I4,5 = 1.2A
- Current across R6: I6 = 10A
Please note that the negative signs indicate the direction of current is opposite to the clockwise direction in the circuit you described.