Question

In a parallel circuit, et = 120 v, r = 3.3 kΩ, and xl = 4 kΩ. What is the value of ir?

Answer 1

To find the current Ir in a parallel circuit with et = 120 V, a resistance of 3.3 kΩ, and an inductive reactance of 4 kΩ, calculate the total impedance and apply Ohm's law to get Ir ≈ 0.02315 A.

Step-by-step explanation:

To calculate the value of Ir in a parallel circuit with an EMF of et = 120 V, a resistance R of 3.3 kΩ, and an inductive reactance XL of 4 kΩ, we use the AC version of Ohm's Law. Given that Ir is the current through the resistor, and the total impedance Z of the circuit is a combination of resistance and reactance, we first must find the total impedance before applying Ohm's law.

The total impedance Z in a parallel circuit having resistance R and reactance XL is given by:

• Z = √(R^2 + XL^2)

For the given values:

• R = 3.3 kΩ = 3300 Ω
• XL = 4 kΩ = 4000 Ω

We now calculate total impedance:

• Z = √(3300^2 + 4000^2) ≈ √(10890000 + 16000000) ≈ √26890000 ≈ 5185.68 Ω

Now, using the AC Ohm's Law Ir = V/Z, where V is the voltage across the circuit, we can find the current Ir:

• Ir = 120 V / 5185.68 Ω ≈ 0.02315 A

This value, 0.02315 A, is the current that flows through the resistor in the circuit.