how do i answer this. A series RL circuit contains two resistors and two inductors.the resistors are 33 ohms and 47 ohms the inductors have inductive reactances of 60 ohms and 30 ohms. the applied voltage is 120 v. what is the voltage drop on the 33 ohm resister?
Z = R + j X
R = 80 Ohms
X = 90 Ohms
|Z|^2 = 80^2 + 90^2
|Z| = 120 ohms
so i = E/Z = 120/120 = 1 amp
V = i R = 1*33 = 33 V
Well, let me enlighten you with a little joke before we dive into the answer:
Why don't scientists trust atoms?
Because they make up everything!
Now, back to your question about the voltage drop in the series RL circuit. We can calculate it by finding the total impedance first. The impedance (Z) of an RL circuit is calculated by adding the resistive and reactive components.
For the inductor with an inductive reactance of 60 ohms, we have Z1 = 60 ohms.
For the inductor with an inductive reactance of 30 ohms, we have Z2 = 30 ohms.
For the resistor with a resistance of 33 ohms, we have R1 = 33 ohms.
For the resistor with a resistance of 47 ohms, we have R2 = 47 ohms.
To find the total impedance (Z), we use the Pythagorean theorem for right triangles:
Z = √(R^2 + (XL - XC)^2)
Where XL and XC are the inductive and capacitive reactances, but since this is an RL circuit, there is no capacitive reactance (XC = 0).
Now, let's calculate:
Z1 = √(33^2 + (60 - 0)^2)
Z2 = √(47^2 + (30 - 0)^2)
Once we have the total impedance (Z), we can use Ohm's Law to calculate the voltage drop (V) across the resistor with a resistance of 33 ohms:
V = (R / Z) * V_applied
Where R is the resistance of the resistor you're interested in, Z is the total impedance, and V_applied is the applied voltage.
Using the values you provided, simply plug in the numbers and crunch the calculations. I hope this helps you on your electrical journey!
To determine the voltage drop on the 33 ohm resistor in the series RL circuit, follow these steps:
Step 1: Calculate the total impedance.
Since the circuit contains both resistors and inductors, we need to find the total impedance (Z) of the circuit. The impedance in a series RL circuit is given by the formula:
Z = √(R^2 + (XL - XC)^2),
where R is the resistance of the circuit, XL is the inductive reactance, and XC is the capacitive reactance.
Given:
Resistance, R1 = 33 ohms
Resistance, R2 = 47 ohms
Inductive reactance, XL1 = 60 ohms
Inductive reactance, XL2 = 30 ohms
To find the total impedance, we use the formula for resistors in series:
RTotal = R1 + R2.
Using the given values, we have:
RTotal = 33 ohms + 47 ohms
RTotal = 80 ohms.
Now let's calculate the inductive reactance and capacitive reactance of the circuit:
XTotal = XL1 + XL2.
Using the given values, we have:
XTotal = 60 ohms + 30 ohms
XTotal = 90 ohms.
Plugging in the values into the formula for total impedance:
Z = √(80^2 + (90)^2)
Z ≈ √(6400 + 8100)
Z ≈ √14500
Z ≈ 120.42 ohms (approximately)
Step 2: Calculate the current in the circuit.
In a series RL circuit, the current (I) is given by the formula:
I = V / Z,
where V is the applied voltage and Z is the total impedance.
Given:
Applied voltage, V = 120 V
Total impedance, Z = 120.42 ohms (approximately)
Plugging in the values into the formula for current:
I = 120 V / 120.42 ohms
I ≈ 0.996 A (approximately)
Step 3: Calculate the voltage drop on the 33 ohm resistor.
The voltage drop across a resistor in a series circuit can be calculated using Ohm's law:
V = I * R,
where V is the voltage drop, I is the current, and R is the resistance of the resistor.
Given:
Resistance, R = 33 ohms
Current, I = 0.996 A (approximately)
Plugging in the values into the formula:
V = 0.996 A * 33 ohms
V ≈ 32.868 V (approximately)
Therefore, the voltage drop on the 33 ohm resistor in the series RL circuit is approximately 32.868 volts.
To find the voltage drop on the 33-ohm resistor in the series RL circuit, you need to first calculate the total impedance of the circuit. Then, using Ohm's Law, you can find the voltage drop across the 33-ohm resistor.
1. Calculate the total impedance (Z) of the circuit:
- Impedance in an RL circuit is given by the formula Z = √(R^2 + (Xl - Xc)^2), where R is the total resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. Since there is no capacitor mentioned in the problem, we can ignore the capacitive reactance, making Xc = 0.
- In this case, the total resistance (R) is the sum of the two resistors: Rtotal = 33 ohms + 47 ohms = 80 ohms.
- The inductive reactances are given as Xl1 = 60 ohms and Xl2 = 30 ohms.
- Substitute the values into the impedance formula: Z = √(80^2 + (60 - 30)^2).
2. Calculate the total impedance:
- Calculate Z: Z = √(6400 + 900) = √(7300) ≈ 85.44 ohms.
3. Calculate the voltage drop on the 33-ohm resistor using Ohm's Law:
- Apply Ohm's Law: V = I * R, where V is the voltage drop, I is the current flowing through the circuit, and R is the resistance.
- Since the circuit is in series, the current flowing through all elements is the same. Let's denote it as I.
- The voltage drop across the 33-ohm resistor is V = I * 33 ohms.
Now, since the current (I) is not specified in the problem, we cannot directly determine the voltage drop on the 33-ohm resistor. The question appears to be incomplete or missing information about the current.