11 LR Transients (68-72)(1)(1)

Switching On an Inductive Circuit

Initial Conditions (t = 0)

• The coil current (i) begins to increase at a maximum rate of change, influenced by the applied voltage (V).

• An induced back electromotive force (emf, e) develops, opposing the change in current as per Lenz’s Law, which states that the direction of the induced emf is such that it opposes the cause producing it.

Current Increase

• As current (i) increases over time, the rate of change of this current begins to decrease, resulting in a lower back emf value.

• The final steady-state coil current (I) is achieved when the back emf (e) finally falls to zero, indicating full current flow through the coil.

Key Equations

• Final current in the coil:[ I = \frac{V}{R_L} \text{ Amperes} ]where ( R_L ) is the total resistance in the coil circuit.

• Current through resistor R1 immediately switches to final value:[ I_1 = \frac{V}{R_1} \text{ Amperes} ]reflecting the instantaneous response of the resistor to the voltage applied.

• Although the total voltage (V) is initially distributed across the coil, the coil current cannot instantaneously rise to its final value (I) due to inductive properties.

Current and Back emf Graph

• The current waveform changes dynamically:

  • At t = 0, the switch is closed, and the current starts at zero.

  • Coil current increases gradually, forming an exponential growth curve until it reaches the maximum value (I) over time.

Time Constant and Switching Off an Inductive Circuit

Time Constant (T)

• The time constant (T) is defined as[ T = \frac{L}{R_L} \text{ seconds} ]where L is the inductance of the coil. This constant indicates how quickly the current in the inductor will reach approximately 63.2% of its final value.

• The coil current then reaches its final value (I) after approximately 5 time constants, signifying a solid approximation for reaching steady-state.

Instantaneous Current & Induced emf

• Coil current at any instant:[ i = I \left(1 - e^{-t/T}\right) \text{ Amperes} ]which describes how the current builds up over time due to the inductance effect.

• Induced emf at any instant:[ e = V e^{-t/T} \text{ Volts} ]showing that the induced emf decreases exponentially as time progresses.

Switching Off

• When the switch (S) is opened, the stored magnetic energy attempts to maintain the current, causing it to flow in the opposite direction through resistor R1 due to the collapsing magnetic field.

• Discharge Path: If there is no adequate discharge path through R1, the back emf generated can cause arcing at the switch contacts, which could lead to equipment damage.

Time constant when switching off:

• The time constant when switching off can be expressed as[ T = \frac{L}{R_1 + R_L} \text{ seconds} ]introducing the load resistances in the formula instead of just the inductance itself.

• Coil current during discharge:[ i = I_0 e^{-t/T} \text{ Amperes} ]where [ I_0 = \frac{V}{R_L} \text{ Amperes} ] , indicating how the current decays over time after the switch is opened.

• Maximum self-induced emf occurs at the very moment of switch-off, highlighting the energy transformation process taking place.

Effects of Opening Switch

• Current and Induced emf Graphs Post-Switch Off:

  • At switch-off, the current through R1 becomes negative, which indicates a reversal of current flow.

  • The last current (I0) just before the switch is opened relates directly to the maximum values obtained right before the opening action.

Self-Assessment Questions on LR Transients

DC Generator Field Coil

• Given: Time constant = 1.5s, Current = 50A, Voltage = 200V, Discharge resistance = 5Ω

• Calculate:a) Inductance of the field.b) Time to drop current to 5A.(Answers: 6H, 1.54s)

Inductance of a Coil

• Given: Inductance = 2H, resistance = 200Ω, voltage = 150V

• Calculate:a) Time constant.b) Coil current after 20ms.c) Maximum induced emf after 0.1s via a 100Ω resistor.d) Current after 15ms switching off.(Answers: 10ms, 648.5mA, 225V, 79mA)

25Ω Resistance Coil

• Given: Resistance = 25Ω, Inductance = 2.5H, Voltage = 50V

• Find:a) Current after 150ms.b) Time to reach current of 1.8A.(Answers: 1.554A, 0.23s)

More Self-Assessment Questions

Non-Inductive Resistor with Coil

• Given: 150Ω non-inductive resistor, 5H inductance, 100Ω resistance, 100V DC supply.

• Calculate:a) Coil current 12ms post switch on.b) Maximum emf induced.c) Coil current 12ms post switch-off.(Answers: 213mA, 175V, 383mA)

Inductance and Resistance Coil

• Given: Inductance = 2H, Resistance = 1Ω, Voltage = 100V

• Calculate:a) Current after 5 seconds.b) Current after disconnecting and switching across a 3Ω resistor.(Answers: 91.79A, 13.53A, 0.5s)

Coil Analysis

• Given: 0.25H inductance, 250Ω resistance, 125V supply

• Calculate:a) Time constant.b) Coil current after 2ms.c) Time for current to reach 0.1A after 4ms off.(Answers: 1ms, 432.3mA, 0.795ms)

Switch Position Effects

• Given: Inductance = 7.5H, Resistance = 75Ω, Voltage = 400V

• Determine:a) Current after 120ms.If switched to position ‘b’ after reaching 90% of final value, find current after 40ms.