Shunt Motor Problems

Study Notes

A shunt motor's supply voltage (V) equals the generated e.m.f. (E) plus the armature volt drop (IaRa).
Alternatively, the generated e.m.f. (E) equals the supply voltage (V) minus the armature volt drop (IaRa).
Supply current (I) equals the armature current (Ia) plus the field current (If).
Example problem: A 240V shunt motor draws a total current of 30A. Rf = 150Ω and Ra = 0.4Ω. Calculate the armature current and the back e.m.f.
- Field current (If) = 240V / 150Ω = 1.6A
- Armature current (Ia) = 30A - 1.6A = 28.4A
- Back e.m.f. (E) = 240V - (28.4A * 0.4Ω) = 228.64V

A 200V DC shunt-wound motor has an armature resistance of 0.4Ω. At a certain load, the armature current is 30A and the motor runs at 1350 rev/min.
If the load is increased to increase the armature current to 45A, determine the motor speed assuming a constant flux.
The relationship E₁/Φ₁n₁ = E₂/Φ₂n₂ applies for both generators and motors, where Φ is constant.
E₁ = 200V - (30A * 0.4Ω) = 188V
E₂ = 200V - (45A * 0.4Ω) = 182V
n₂ = (E₂ * n₁) / E₁ = (182V * 1350rev/min) / 188V = 1307 rev/min

In a series motor, the field winding is in series with the armature.
Supply voltage (V) equals the generated e.m.f. (E) plus the total resistance drop (I * (Ra + Rf))
Alternatively, the generated e.m.f. (E) equals the supply voltage (V) minus the total resistance drop (I * (Ra + Rf))
Example problem: A series motor has an armature resistance of 0.2Ω and a series field resistance of 0.3Ω, connected to a 240V supply. The motor runs at 24 rev/s when drawing 15A.
- Calculate the generated e.m.f. at this load.
- Calculate the speed if the load is increased so the current is 30A and assuming the flux doubles.
  - E₁ = 240V - (15A * (0.2Ω + 0.3Ω)) = 232.5V
  - E₂ = 240V - (30A * (0.5 Ω)) = 225V

When a stationary DC motor is switched directly to the supply, fuses may burn out due to low armature resistance.
To start the motor, additional resistance is added to the armature circuit.
As the motor speed increases, the armature current is limited by a generated voltage that counters the applied voltage.
Additional resistance may be reduced.

This is a method to control the speed of DC motors, particularly in applications needing wide-ranging speed control.
It is a motor-generator set.
Components:
- Prime Mover (AC motor): A constant-speed motor driving the DC generator.
- DC Generator: Converts mechanical energy to electrical energy supplying the DC motor.
- DC Motor: Receives power from the DC generator and drives the mechanical load.
- Control System: Adjusts the field winding of the DC generator, thus regulating the output voltage & subsequently the DC motor speed.

Modern devices used for precise speed control of electric motors.
Advantages: High efficiency, compact size, and precision.
Solid-state controllers have largely replaced traditional methods, like the Ward-Leonard system.
Working Principle: Solid-state controllers adjust motor parameters like voltage, frequency, and current to regulate speed.
Types of Solid-State Speed Controllers:
- Variable Frequency Drives (VFDs): Used for controlling AC motors by varying frequency and voltage of the supply. Primarily used in HVAC, conveyor belts, and pumps
- Soft Starters: Gradually ramp-up motor torque to avoid stress on mechanical components during startup.
- DC Choppers and Phase-Controlled Rectifiers: Used for controlling DC motors by adjusting voltage using time-dependent voltage modulation.

In a separately excited generator, the field winding is excited by a separate DC supply.
The induced e.m.f. (E) = V + IR_a
When a load is connected, a load current (I_a) flows, leading to a voltage drop (I_aR_a) which reduces the terminal voltage (V).