Skip to main content

Long shunt compound generator emf, 40 A at 200 V

Short answer

I_sh=200/100=2 A, I_a=42 A. E=200+42×(0.4+0.5)=237.8 V.

A dc long shunt compound generator has armature winding resistance of 0.4 ohm, series field winding resistance of 0.5 ohm and shunt-field winding resistance of 100 ohms. The generator delivers a current of 40 A to the load at 200 volt. Calculate the emf generated by the armature.

In a long shunt compound generator the shunt field is connected across the load terminals, so the series field carries the full armature current.

Given: IL=40I_L = 40 A, V=200V = 200 V, Rsh=100 ΩR_{sh} = 100\ \Omega, Rse=0.5 ΩR_{se} = 0.5\ \Omega, Ra=0.4 ΩR_a = 0.4\ \Omega.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Step 1: Shunt field current

Ish=VRsh=200100=2.00 AI_{sh} = \frac{V}{R_{sh}} = \frac{200}{100} = 2.00\ \text{A}

Step 2: Armature current

Ia=Ise=IL+Ish=40+2.00=42.00 AI_a = I_{se} = I_L + I_{sh} = 40 + 2.00 = 42.00\ \text{A}

Step 3: Generated emf

E=V+Ia(Ra+Rse)=200+42.00(0.4+0.5)=200+37.800=237.80 VE = V + I_a(R_a + R_{se}) = 200 + 42.00(0.4 + 0.5) = 200 + 37.800 = 237.80\ \text{V}

Answer: emf generated by the armature E=237.80 VE = 237.80\ \text{V}.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines chapter 3 questions →