**The field winding of a shunt DC (Direct Current) motor has a resistance of 110 ohms, and the emf (electro-motive force) applied to it is 220 volts. What is the amount of power expended in the field excitation?**

__Example 1__:

**If***=*

*Et*

**÷**

*Rf**- field current*

**If***- EMF armature terminal*

**Et***- field resistance*

**Rf***= 220 volts ÷ 110 ohms*

**If***= 2 amperes*

**If***x*

**Et***= 220 volts x 2 amperes = 440 watts.*

**If****=**

*Pf*

*Et***÷**

*2***= (220)2 ÷ 110 = 440 watts**

*Rf***A shunt DC motor whose armature resistance is 0.2 ohm and whose terminal voltage is 220 volts requires an armature current of 50 amperes and runs at 1,500 RPM when the field is fully excited. If the strength of the field is decreased and the amount of armature current is increased, both by 50%, at what speed will the motor run?**

__Example 2__:**=**

*Ea***-**

*Et***x**

*Ia*

*Ra***= 220 volts - (50 amperes x 0.2 ohm) = 210 volts**

*Ea1***= 220 volts - (75 amperes x 0.2 ohm) = 205 volts**

*Ea2***=**

*Ea*

*N*

*ɸ*

*K***÷**

*Ea1***=**

*Ea2***÷**

*N1 ɸ1 K1*

*N2 ɸ2 K2***= 1.5**

*ɸ1***, and**

*ɸ2***=**

*Z1*

*Z2*

*N2**N2 =*1,500 x 205 x 1.5 ÷ 210 = 2,196 RPM

**: A 7.5 horsepower 220 volts interpole DC motor has armature and shunt field resistances of 0.5 ohm and 200 ohms respectively. The current input at 1,800 RPM under no-load conditions is 3.5 amperes. What are the current and electromagnetic torque for a speed of 1,700 RPM?**

__Example 3__**=**

*Ia***-**

*I*_{L}**= 3.5 amperes - (220 volts ÷ 200 ohms) = 2.4 amperes**

*If*

*ɸ K*

*NL**= Et - (Ia Ra)*

*÷**N =*220 - (2.4 x 0.5)

**÷**1,800 = 0.1216

*ɸ K***=**

*NL*

*ɸ KFL***=**

*Ia***- (**

*Et***) ÷**

*N ɸ K***= 220 - (1,700 x 0.1216) ÷ 0.5 = 26.6 amperes**

*Ra***=**

*I*_{L}**+**

*Ia***= 26.6 + 1.1 = 27.7 amperes**

*If***(electromagnetic torque) = 7.05**

*Te*

*K*

*ɸ***= 7.05 x 0.1216 x 26.6 = 22.8 ft-lb.**

*Ia***: The mechanical efficiency of a shunt DC motor whose armature and field resistances are 0.055 and 32 ohms respectively, is to be tested by means of a rope brake. When running at 1,400 RPM, the longitudinal pull on the 6 inch diameter pulley is 57 lbs. Simultaneous readings on the line voltmeter and ammeter are 105 volts and 35 amperes respectively. Calculate (a) the counter emf developed; (b) the copper losses; (c) the efficiency.**

__Example 4__**=**

*Ia*

*I***-**

_{L}**= 35 - (105/32) = 31.7 amperes**

*If***=**

*Ea***- (**

*Et***x**

*Ia***) = 105 - (31.7 x 0.055) = 103.26 volts**

*Ra***= (**

*Pc*

*If***x**

*2***) + (**

*Rf*

*Ia***x**

*2***) = [(3.3)2 x (32)] + [(31.7)2 x (0.055)] = 404 watts**

*Ra*