Skip to main content

EMF of short shunt compound generator (50 A, 220 V)

Short answer

Shunt field voltage is 222.5 V, shunt current 2.225 A, armature current 52.225 A. Generated emf is 227.72 V.

A short shunt dc compound generator supplies a current of 50 A at 220 V. The shunt field, series field and armature winding resistances are 100 ohm, 0.05 ohm and 0.1 ohm respectively. Calculate the emf generated by the armature.

Given: IL=50I_L = 50 A, V=220V = 220 V, Rsh=100 ΩR_{sh} = 100\ \Omega, Rse=0.05 ΩR_{se} = 0.05\ \Omega, Ra=0.1 ΩR_a = 0.1\ \Omega; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 220 V
     |      |                 |
     +------+-----------------+
Vsh=V+ILRse=220+50×0.05=222.5 VIsh=222.5100=2.225 AIa=IL+Ish=50+2.225=52.225 AEg=Vsh+IaRa=222.5+52.225×0.1=222.5+5.22=227.72 V\begin{aligned} V_{sh} &= V + I_L R_{se} = 220 + 50 \times 0.05 = 222.5\ \text{V} \\ I_{sh} &= \frac{222.5}{100} = 2.225\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 2.225 = 52.225\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 222.5 + 52.225 \times 0.1 \\ &= 222.5 + 5.22 = 227.72\ \text{V} \end{aligned}

Brush drop is neglected as it is not given.

Answer: EMF generated Eg≈227.72E_g \approx 227.72 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines I chapter 3 questions →