Math Problem Statement

The current in the RL circuit shown below is i(t) = 10 cos(5t - π/4) A. Find the value of the inductor?

Solution

We are tasked with finding the value of the inductor LL in an RL circuit based on the provided current equation i(t)=10cos(5tπ/4)i(t) = 10 \cos(5t - \pi/4) A and the voltage source v(t)=200cos(5t)v(t) = 200 \cos(5t) V.

Step 1: Impedance of the Circuit

In an RL circuit, the impedance ZRLZ_{RL} is given by: ZRL=R+jωLZ_{RL} = R + j\omega L where RR is the resistance, LL is the inductance, and ω=5rad/s\omega = 5 \, \text{rad/s} (since the frequency of the voltage is 5t5t).

Step 2: Phasor Form

To analyze the circuit in phasor form:

  • Voltage: V(t)=200cos(5t)V(t) = 200 \cos(5t) corresponds to phasor V=2000V = 200 \angle 0^\circ.
  • Current: i(t)=10cos(5tπ/4)i(t) = 10 \cos(5t - \pi/4) corresponds to phasor I=1045I = 10 \angle -45^\circ.

Step 3: Applying Ohm's Law in Phasor Domain

Using Ohm's law for phasors, the total impedance ZRLZ_{RL} is given by: ZRL=VIZ_{RL} = \frac{V}{I} Substituting the known phasors: ZRL=20001045=2045ΩZ_{RL} = \frac{200 \angle 0^\circ}{10 \angle -45^\circ} = 20 \angle 45^\circ \, \Omega This means that the impedance has a magnitude of 20 ohms and a phase angle of 4545^\circ.

Step 4: Expressing Impedance in Rectangular Form

The impedance ZRL=2045Z_{RL} = 20 \angle 45^\circ can be expressed in rectangular form as: ZRL=20(cos45+jsin45)=20(12+j12)Z_{RL} = 20 (\cos 45^\circ + j \sin 45^\circ) = 20 \left( \frac{1}{\sqrt{2}} + j \frac{1}{\sqrt{2}} \right) Simplifying: ZRL=20×12(1+j)=14.14+j14.14ΩZ_{RL} = 20 \times \frac{1}{\sqrt{2}} (1 + j) = 14.14 + j14.14 \, \Omega

Step 5: Identifying the Resistance and Inductance

From the impedance, we identify:

  • The real part, R=14.14ΩR = 14.14 \, \Omega, which represents the resistance.
  • The imaginary part, ωL=14.14Ω\omega L = 14.14 \, \Omega.

Step 6: Solving for Inductance LL

Since ω=5rad/s\omega = 5 \, \text{rad/s}, we can solve for LL: ωL=14.14    L=14.145=2.828H\omega L = 14.14 \implies L = \frac{14.14}{5} = 2.828 \, \text{H}

Final Answer:

The inductance LL is 2.828H2.828 \, \text{H}.

Let me know if you'd like further details or have questions!

Here are 5 related questions to expand your understanding:

  1. How would the value of LL change if the frequency of the voltage source increased?
  2. Can you derive the time-domain voltage across the inductor using the current i(t)i(t)?
  3. What would happen to the current phase angle if RR increased?
  4. How would the impedance of the circuit change if the frequency doubled?
  5. What is the relationship between voltage and current in a purely inductive circuit?

Tip: When analyzing AC circuits, converting to the phasor domain simplifies solving for impedance and current relationships.

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Math Problem Analysis

Mathematical Concepts

AC Circuit Analysis
Phasor Analysis
Impedance in RL Circuits

Formulas

Impedance of RL Circuit: Z_{RL} = R + jωL
Phasor Ohm's Law: Z = V/I
Rectangular to Polar Conversion: Z = R + jX = |Z|∠θ

Theorems

Ohm's Law in Phasor Form
AC Circuit Impedance

Suitable Grade Level

Undergraduate Level