RL Circuit Analysis Exercise

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Questions and Answers

What is the inductance L of the coil in Exercise 01?

  • 200 mH
  • 68 mH
  • 75 mH (correct)
  • 12 mH

Which of the following equations represents the voltage across the resistor U_R(t) in the circuit?

  • $U_R(t) = rac{E}{R}(1-e^{- rac{R}{L}t})$
  • $U_R(t) = rac{E}{R+r}(1 + e^{- rac{R}{L}t})$
  • $U_R(t) = rac{R.E}{R+r} (1- e^{- rac{R+r}{L}t})$ (correct)
  • $U_R(t) = E - U_L(t)$

What is the expression for the maximum current I in the circuit in Exercise 01?

  • $I = R / E$
  • $I = \frac{E}{R+r}$ (correct)
  • $I = E / R$
  • $I = \frac{R+r}{E}$

What does the time constant Ï„ of the RL circuit represent?

<p>The time it takes for the current to reach 63.2% of its maximum value (C)</p> Signup and view all the answers

Which statement about the differential equation for the voltage across the resistor is true?

<p>It includes both the resistance of the resistor and the internal resistance of the coil. (B)</p> Signup and view all the answers

When observing the curves on the oscilloscope, Curve 1 represents which parameter?

<p>Voltage across the resistor (B)</p> Signup and view all the answers

What might be a plausible reason for the internal resistance r to be significant in the RL circuit?

<p>It contributes to energy loss in the form of heat. (D)</p> Signup and view all the answers

How is the energy stored in the coil related to the inductance L and the maximum current I?

<p>$Energy = \frac{1}{2} L I^2$ (C)</p> Signup and view all the answers

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Study Notes

Exercise 1: RL Circuit Analysis

  • Circuit configuration: A coil with inductance (L = 75 mH) and internal resistance (r = 11.5 Ω) is connected in series with a resistor (R = 68 Ω) and a constant voltage generator (E = 12 V).
  • Differential equation: The differential equation governing the current (i(t)) in the coil is: L * di/dt + (R+r) * i(t) = E
  • Solution form: The solution to the differential equation takes the form: i(t) = A + B*e^(-(R+r)/L)*t
  • Maximum current: The maximum current (I) in the circuit is given by: I = E / (R+r)
  • Voltage across the coil: The voltage across the coil (UL(t)) is expressed as: UL(t) = E / (R+r) * (r + R * e^(-(R+r)/L)*t)
  • Voltage across the resistor: The voltage across the resistor (UR(t)) is expressed as: UR(t) = R*E / (R+r) * (1 - e^(-(R+r)/L)*t)
  • Energy stored in the coil: The energy stored in the coil in steady state is given by: W = (1/2) * L * I^2

Exercise 2: RL Circuit Analysis with Oscilloscope Readings

  • Circuit configuration: A coil with inductance (L) and internal resistance (r) is connected in series with a resistor (R = 200 Ω) and a constant voltage generator (Uo).
  • Oscilloscope readings: Channel 1 traces the voltage across resistor (R), Channel 2 traces the voltage across the coil.
  • Differential equation for UR(t): (L/(R+r)) * dUR/dt + UR(t) = (RE)/(R+r)
  • Curve 1 analysis: Curve 1 in Figure 1 represents UR(t), the voltage across the resistor.
  • Generator voltage: The generator voltage (Uo) can be determined by analyzing the steady-state value of UR(t) in Figure 1.
  • Time constant: The time constant (Ï„) of the RL circuit is determined by the time it takes for UR(t) to reach approximately 63.2% of its final value in Figure 1.
  • Steady-state current: The steady-state current (I) in the RL circuit is calculated as I = Uo / (R+r)
  • Inductance: The inductance (L) of the coil is calculated using the time constant: L = Ï„ * (R+r)
  • Internal resistance: The internal resistance (r) of the coil is calculated using the steady-state current and the generator voltage: r = Uo/I - R

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