The four quadrant separately excited by a DC motor shown in the figure is powered from a 100 V battery. The field current is kept constant. The armature resistance is 0.6 Ω. When the converter is operating at a duty cycle of 0.8 and armature is drawing constant current of 30 A with negligible ripple, what is the back emf developed across the coil?

Explanation:

Step-by-Step Solution:

1. Understanding the Duty Cycle:
   • The duty cycle is the fraction of time the converter conducts during one period. A duty cycle of 0.8 implies that the converter is ON for 80% of the time and OFF for the remaining 20%.
   • The equivalent voltage applied to the armature can be calculated using the duty cycle and battery voltage.
2. Voltage Applied to the Armature:
   • The converter modifies the battery voltage depending on the duty cycle. The applied voltage (V_applied) is given by:
   • V_applied = Duty Cycle × Battery Voltage
   • Substituting the given values:
   • V_applied = 0.8 × 100 V = 80 V
3. Armature Voltage Equation:
   • The voltage applied to the armature is distributed across the armature resistance and the back emf (E_b). The armature voltage equation is:
   • V_applied = E_b + I_a × R_a
   • Where:
     • E_b = Back emf developed across the motor
     • I_a = Armature current (30 A)
     • R_a = Armature resistance (0.6 Ω)
4. Substitute the Known Values:
   • From the equation:
   • 80 V = E_b + (30 A × 0.6 Ω)
   • Simplifying the resistance term:
   • 30 A × 0.6 Ω = 18 V
   • Therefore:
   • 80 V = E_b + 18 V
   • Solve for E_b:
   • E_b = 80 V - 18 V = 62 V

Back emf developed across the coil:

The back emf across the coil is calculated to be 62 V