a series RL circuit contains two resistors and two inductors. the resistors are 33 ohms and 47 ohms. inductors have inductive reactances of 60 ohms and 30 ohms. the applied voltage is 120 volts. what is the voltage drop on the 33 ohm resistor?
Impedance=sqrt (80^2+90^2)
current=120/impedance
voltage33=33(current)
How would you solve for the 30 ohm inductor?
To find the voltage drop on the 33-ohm resistor in a series RL circuit, we need to calculate the total impedance and then use Ohm's Law.
1. Impedance (Z) is the combination of resistance (R) and inductive reactance (XL) in a series RL circuit. The formula for calculating impedance is:
Z = √(R^2 + XL^2)
Given that R1 = 33 ohms and XL1 = 60 ohms, we can calculate the impedance for the first inductor-resistor combination:
Z1 = √(33^2 + 60^2)
= √(1089 + 3600)
= √4689
≈ 68.43 ohms
2. Following the same process, we can calculate the impedance for the second inductor-resistor combination:
Z2 = √(47^2 + 30^2)
= √(2209 + 900)
= √3109
≈ 55.76 ohms
3. The total impedance (Zt) of the series RL circuit is the sum of the individual impedances:
Zt = Z1 + Z2
≈ 68.43 ohms + 55.76 ohms
≈ 124.19 ohms
4. Once we have the total impedance, we can apply Ohm's Law to calculate the voltage drop on the 33-ohm resistor, which is the same as the across the first inductor-resistor combination:
V1 = I * Z1
Since the circuit is in series, the current flowing through the circuit (I) is the same as the current flowing through each component. To calculate the current, we can use Ohm's Law:
Vt = I * Zt
I = Vt / Zt
Given that the applied voltage (Vt) is 120 volts, we can calculate the current:
I = 120 volts / 124.19 ohms
≈ 0.967 A
Now, we can calculate the voltage drop on the 33-ohm resistor:
V1 = 0.967 A * 68.43 ohms
≈ 66.26 volts
Therefore, the voltage drop on the 33-ohm resistor is approximately 66.26 volts.
To find the voltage drop across the 33-ohm resistor in a series RL circuit, you need to calculate the total impedance of the circuit first, and then use Ohm's law to determine the voltage drop across the resistor.
Here's how you can calculate it step by step:
Step 1: Calculate the total impedance (Z) of the circuit.
- Impedance is the total opposition to current flow in an AC circuit and is represented by a complex number.
- In a series RL circuit, the total impedance is the phasor sum of the resistive and inductive reactance components.
- The resistive component is the sum of the resistances, and the inductive reactance component is the sum of the inductive reactances.
- Therefore, the total impedance (Z) can be calculated as Z = R + jXL, where R is the sum of the resistances, XL is the sum of the inductive reactances, and j is the imaginary unit.
- In this case, R = 33 ohms + 47 ohms = 80 ohms, and XL = 60 ohms + 30 ohms = 90 ohms. So, Z = 80 + j90 ohms.
Step 2: Calculate the current flowing through the circuit.
- The current (I) can be calculated using Ohm's law: I = V/Z, where V is the applied voltage and Z is the total impedance.
- In this case, V = 120 volts, and Z = 80 + j90 ohms (from step 1).
- So, I = 120 volts / (80 + j90 ohms).
Step 3: Calculate the voltage drop across the 33-ohm resistor.
- Since the circuit is in series, the voltage drop across each component is the same as the applied voltage.
- Therefore, the voltage drop across the 33-ohm resistor is equal to the applied voltage, which is V = 120 volts.
Thus, the voltage drop on the 33-ohm resistor in the series RL circuit is 120 volts.