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 there is a connection of point A to point B. In between there is a parallel circuit with R2= 10 ohms to R3=2ohms on the top. On the bottom there is another parallel circuit consisting of R4= 20 ohms and R5= 5 ohms. After point B is one more resistor R6= 6 ohms. Find the voltage drop and current across each resistor.
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?
I don't understand from your description where points A and B are.
The two parallel resistors of 10 and 2 ohms are equivelant to 1 2/3 ohms in series. The two parallel resistors of 20 and 5 ohms are equivalent to 4 ohms in series.
Total resistance that the battery see is
3 + 1 2/3 + 4 + 6 ohms = 14 2/3
Use this and ohm's law to get the current in the series parts of the circuit.
The current flowing through an electrical circuit is indirectly proportional to the resistance in the circuit. If a current of 8A flows through the circuit when the resistance is 30 ohms, how much current will flow through the circuit if the resistance is 14 ohms. show your answer to 3 significant figures.
To solve this problem, we'll start by determining the total resistance of the parallel circuits at points A and B.
For the top parallel circuit (R2 = 10 ohms and R3 = 2 ohms), we can calculate the total resistance (Rt) using the formula:
1/Rt = 1/R2 + 1/R3
1/Rt = 1/10 + 1/2
1/Rt = 1/10 + 5/10
1/Rt = 6/10
1/Rt = 3/5
Rt = 5/3 ohms
For the bottom parallel circuit (R4 = 20 ohms and R5 = 5 ohms), we can calculate the total resistance (Rb) using the same formula:
1/Rb = 1/R4 + 1/R5
1/Rb = 1/20 + 1/5
1/Rb = 1/20 + 4/20
1/Rb = 5/20
1/Rb = 1/4
Rb = 4 ohms
Now, we'll determine the total resistance (Rt2) for the series circuit formed by R6 (6 ohms) and Rt (5/3 ohms). For series circuits, we simply add the resistances:
Rt2 = R6 + Rt
Rt2 = 6 + 5/3
Rt2 = 18/3 + 5/3
Rt2 = 23/3 ohms
Next, we'll find the current (I) flowing through the circuit using Ohm's law, V = IR. We know the total voltage (V) is 120V, and we'll use Rt2 as the total resistance:
I = V / Rt2
I = 120 / (23/3)
I = 360/23 A
Now we can find the voltage drop (V1) and current (I1) across R1 (3 ohms) using Ohm's law:
V1 = I1 * R1
I1 = V1 / R1
Since R1 is in series with the parallel circuit, the current flowing through it is the same as the total current (I) in the circuit:
I1 = I
I1 = 360/23 A
V1 = (360/23) * 3
V1 = 1080/23 V
For the other resistors, we can use the fact that the voltage across parallel branches in a circuit is the same.
The voltage across R2 (10 ohms) and R3 (2 ohms) will be equal to R1 (V1), which is 1080/23V.
Similarly, the voltage across R4 (20 ohms) and R5 (5 ohms) will also be equal to V1, which is 1080/23V.
Therefore, the voltage drop across R2, R3, R4, and R5 is 1080/23V each.
Now we can calculate the current (I2) flowing through R2 and R3, and the current (I3) flowing through R4 and R5, using Ohm's law:
I2 = V1 / R2
I2 = (1080/23) / 10
I2 = 1080/230 A
I3 = V1 / R4
I3 = (1080/23) / 20
I3 = 1080/460 A
Finally, we know the current (I) flowing through R6 is the same as the total current in the circuit, 360/23 A.
I hope this helps you solve the problem and find the voltage drop and current across each resistor. Let me know if you have any further questions!