DC Motors

LessonAbout 1.5 h

DC motor is an electric motor that converts direct-current electrical energy into mechanical rotational energy by electromagnetic interaction.

1. Speed Characteristics of a DC Motor

Speed characteristic defines the relationship between armature speed and load current or magnetic flux in a DC motor.

Speed is directly proportional to back emf
Speed is inversely proportional to magnetic flux per pole
Armature reaction weakens magnetic flux under heavy loads

DC Motor Speed Equations

\[ N = \dfrac{60 a E_b}{P Z \Phi} \]

\[ N = k \dfrac{E_b}{\Phi} \]

Where: \(N\) = rotational speed (rpm), \(E_b\) = back emf (V), \(a\) = parallel current paths, \(P\) = number of poles, \(Z\) = total armature conductors, \(\Phi\) = flux per pole (Wb), \(k\) = motor speed constant

Example. At rated load, a shunt motor draws an armature current of \(50\ \text{A}\) from a \(230\ \text{V}\) DC mains while running at \(1000\ \text{rpm}\). At no load the armature current drops to \(5\ \text{A}\). If the effect of armature reaction weakens the flux by \(2\%\) from no-load to full-load, determine the speed of the motor at no-load. Assume armature resistance to be \(0.15\ \Omega\).

Solution.

\[ E_{b1} = 230 - 50(0.15) = 222.5\ \text{V} \qquad\qquad E_{b2} = 230 - 5(0.15) = 229.25\ \text{V} \]

\[ N_2 = N_1 \left(\dfrac{E_{b2}}{E_{b1}}\right)\left(\dfrac{\Phi_1}{\Phi_2}\right) = 1000\left(\dfrac{229.25}{222.5}\right)(0.98) = 1010\ \text{rpm} \]

Answer: \(1010\ \text{rpm}\)

2. Torque Developed in the Armature

Armature torque is the total electromagnetic turning moment developed inside the rotating armature core.

Proportional to product of flux per pole and armature current
Independent of motor friction and mechanical losses
Converted into mechanical power in the rotating armature

Armature Torque Formulas

\[ T = \dfrac{P Z \Phi I_a}{2 \pi a} \]

\[ T = 9.55 \left(\dfrac{P_d}{N}\right) \]

\[ P_d = E_b I_a \]

Where: \(T\) = torque developed (\(\text{N}\cdot\text{m}\)), \(P_d\) = electromagnetic power developed (W), \(N\) = speed of rotation (rpm), \(E_b\) = back emf (V), \(I_a\) = armature current (A), \(P\) = number of poles, \(Z\) = total armature conductors, \(\Phi\) = flux per pole (Wb), \(a\) = parallel current paths

Example. A 4-pole DC motor has a lap-wound armature with \(580\) conductors. The flux per pole is \(20\ \text{mWb}\) and the armature current is \(125\ \text{A}\). Determine the electromagnetic torque developed in \(\text{N}\cdot\text{m}\).

Solution.

\[ T = \dfrac{Z \Phi I_a}{2 \pi m} = \dfrac{580(0.020)(125)}{2 \pi (1)} = 230.77\ \text{N}\cdot\text{m} \]

Answer: \(230.77\ \text{N}\cdot\text{m}\)

3. Mechanical Power Output

Mechanical power output is the net shaft horsepower delivered to drive external mechanical equipment.

Equal to total developed power minus rotational losses
Expressed in mechanical horsepower or kilowatts
Evaluated from shaft output torque and operating speed

Mechanical Power Output Formulas

\[ \text{Hp} = \dfrac{2 \pi N T}{33000} \]

\[ \text{Hp} = \dfrac{2 \pi N T}{44760} \]

Where: \(\text{Hp}\) = output power (hp), \(N\) = shaft speed (rpm), \(T\) = shaft torque (\(\text{lb}\cdot\text{ft}\) for 33,000; \(\text{N}\cdot\text{m}\) for 44,760)

Example. A 4-pole, \(500\ \text{V}\) DC motor has a lap-connected armature with \(870\) conductors. The flux per pole is \(0.06\ \text{Wb}\) and the armature draws \(300\ \text{A}\). If the mechanical efficiency is \(96\%\), determine the brake output horsepower.

Solution.

\[ E_b = \dfrac{P N Z \Phi}{60 a} = \dfrac{500(870)(0.06)}{60(1)} = 435\ \text{V} \]

\[ P_d = \dfrac{E_b I_a}{746} = \dfrac{435(300)}{746} = 175\ \text{hp} \]

\[ P_{\text{brake}} = 0.96(175) = 168\ \text{hp} \]

Answer: \(168\ \text{hp}\)

4. Shunt Motor

Shunt motor has its field winding connected in parallel across the armature circuit terminals.

Operates at nearly constant speed from no-load to full-load
Field flux remains virtually constant during normal operation
Armature current varies directly with mechanical load

Shunt Motor Formulas

\[ I_{\text{sh}} = \dfrac{V_s}{R_{\text{sh}}} \]

\[ I_m = I_a + I_{\text{sh}} \]

\[ E_b = V_s - I_a R_a' \]

Where: \(I_{\text{sh}}\) = shunt field current (A), \(V_s\) = supply voltage (V), \(R_{\text{sh}}\) = shunt field resistance (\(\Omega\)), \(I_m\) = line current (A), \(I_a\) = armature current (A), \(E_b\) = back emf (V), \(R_a'\) = armature circuit resistance (\(\Omega\))

EE Board Exam April 1990

Example. A shunt motor draws a current of \(40\ \text{A}\) from a \(120\ \text{V}\) source and runs at \(1200\ \text{rpm}\) at rated load. The armature and field circuit resistances are \(0.1\ \Omega\) and \(60\ \Omega\) respectively. Determine the speed of the motor at half load. Assume a brush drop of \(2\ \text{V}\) at rated load and \(1\ \text{V}\) at half load.

Solution.

\[ I_{\text{sh}} = \dfrac{120}{60} = 2\ \text{A} \qquad\qquad I_{a1} = 40 - 2 = 38\ \text{A} \]

\[ E_{b1} = 120 - 38(0.1) - 2 = 114.2\ \text{V} \qquad\qquad I_{a2} = \dfrac{38}{2} = 19\ \text{A} \]

\[ E_{b2} = 120 - 19(0.1) - 1 = 117.1\ \text{V} \]

\[ N_2 = 1200 \left(\dfrac{117.1}{114.2}\right) = 1230\ \text{rpm} \]

Answer: \(1230\ \text{rpm}\)

5. Series Motor

Series motor has its field winding connected in series with the armature winding and supply line.

Develops high starting torque proportional to current squared
Field flux is directly proportional to armature current
Must never be operated under no-load condition

Series Motor Formulas

\[ I_a = I_{\text{se}} = I_m \]

\[ E_b = V_s - I_a(R_a' + R_{\text{se}}) \]

Where: \(I_a\) = armature current (A), \(I_{\text{se}}\) = series field current (A), \(I_m\) = line current (A), \(E_b\) = back emf (V), \(V_s\) = supply voltage (V), \(R_a'\) = armature resistance (\(\Omega\)), \(R_{\text{se}}\) = series field resistance (\(\Omega\))

Example. A \(110\ \text{V}\) series motor takes \(20\ \text{A}\) while running at \(1000\ \text{rpm}\). Total armature and series field resistance is \(0.1\ \Omega\). Determine the speed if the line current increases to \(50\ \text{A}\) and the flux increases by \(10\%\).

Solution.

\[ E_{b1} = 110 - 20(0.1) = 108\ \text{V} \qquad\qquad E_{b2} = 110 - 50(0.1) = 105\ \text{V} \]

\[ N_2 = N_1 \left(\dfrac{E_{b2}}{E_{b1}}\right) \left(\dfrac{\Phi_1}{\Phi_2}\right) = 1000 \left(\dfrac{105}{108}\right) \left(\dfrac{1}{1.10}\right) = 884\ \text{rpm} \]

Answer: \(884\ \text{rpm}\)

6. Long-Shunt Compound Motor

Long-shunt compound motor has its shunt field winding connected across the series combination of armature and series field.

Combines high starting torque with stable speed regulation
Series field current equals armature current
Cumulative compounding aids shunt field flux

Long-Shunt Compound Motor Formulas

\[ I_{\text{se}} = I_a \]

\[ I_m = I_a + I_{\text{sh}} \]

\[ I_{\text{sh}} = \dfrac{V_s}{R_{\text{sh}}} \]

\[ E_b = V_s - I_a(R_a' + R_{\text{se}}) \]

Where: \(I_{\text{se}}\) = series field current (A), \(I_a\) = armature current (A), \(I_m\) = supply current (A), \(I_{\text{sh}}\) = shunt field current (A), \(V_s\) = supply voltage (V), \(R_{\text{sh}}\) = shunt field resistance (\(\Omega\)), \(E_b\) = back emf (V), \(R_a'\) = armature resistance (\(\Omega\)), \(R_{\text{se}}\) = series field resistance (\(\Omega\))

Example. A \(5.5\ \text{kW}, 125\ \text{V}\) long-shunt compound motor draws \(44\ \text{A}\) load current at rated load. The shunt field resistance is \(50\ \Omega\), armature resistance is \(0.2\ \Omega\), series field resistance is \(0.25\ \Omega\), and stray power losses are \(336.25\ \text{W}\). Solve for the overall efficiency.

Solution.

\[ I_{\text{sh}} = \dfrac{125}{50} = 2.5\ \text{A} \qquad\qquad I_a = 44 + 2.5 = 46.5\ \text{A} \]

\[ P_a = (46.5)^2(0.2) = 432.45\ \text{W} \qquad\qquad P_{\text{sh}} = 125(2.5) = 312.5\ \text{W} \]

\[ P_{\text{se}} = (46.5)^2(0.25) = 540.56\ \text{W} \]

\[ \eta = \dfrac{5500}{5500 + 432.45 + 312.5 + 540.56 + 336.25} \times 100\% = 83.57\% \]

Answer: \(83.57\%\)

7. Short-Shunt Compound Motor

Short-shunt compound motor has its shunt field winding connected directly in parallel across the armature terminals only.

Series field winding carries total supply line current
Shunt field voltage equals armature terminal voltage drop
Differential compounding reduces flux under heavy loads

Short-Shunt Compound Motor Formulas

\[ I_{\text{se}} = I_m \]

\[ I_m = I_a + I_{\text{sh}} \]

\[ I_{\text{sh}} = \dfrac{V_s - I_{\text{se}} R_{\text{se}}}{R_{\text{sh}}} \]

\[ E_b = V_s - I_{\text{se}} R_{\text{se}} - I_a R_a' \]

Where: \(I_{\text{se}}\) = series field current (A), \(I_m\) = supply current (A), \(I_a\) = armature current (A), \(I_{\text{sh}}\) = shunt field current (A), \(V_s\) = supply voltage (V), \(R_{\text{se}}\) = series field resistance (\(\Omega\)), \(R_{\text{sh}}\) = shunt field resistance (\(\Omega\)), \(E_b\) = back emf (V), \(R_a'\) = armature resistance (\(\Omega\))

Example. A \(230\ \text{V}\) short-shunt compound motor takes \(42\ \text{A}\) line current. The resistances of armature, series field, and shunt field are \(0.1\ \Omega\), \(0.2\ \Omega\), and \(50\ \Omega\) respectively. Stray power losses are \(400\ \text{W}\). Calculate the efficiency.

Solution.

\[ I_{\text{sh}} = \dfrac{230}{50} = 4.6\ \text{A} \qquad\qquad I_a = 42 - 4.6 = 37.4\ \text{A} \]

\[ P_a = (37.4)^2(0.1) = 139.876\ \text{W} \qquad\qquad P_{\text{se}} = (37.4)^2(0.2) = 279.752\ \text{W} \]

\[ P_{\text{sh}} = 4.6(230) = 1058\ \text{W} \]

\[ P_{\text{losses}} = 139.876 + 279.752 + 1058 + 400 = 1877.628\ \text{W} \]

\[ P_{\text{in}} = 230(42) = 9660\ \text{W} \qquad\qquad \eta = \dfrac{9660 - 1877.628}{9660} \times 100\% = 80.56\% \]

Answer: \(80.56\%\)

8. Motor Starting

Motor starting is the initial acceleration process when a stationary DC motor is connected to the supply lines.

8.1 Without Rheostat

Starting without rheostat applies full line voltage directly to the stationary armature circuit.

Initial back emf is zero at standstill (\(E_b = 0\))
Starting armature current is extremely high
May damage commutator bars or trip protection breakers

Armature Starting Current Without Rheostat

\[ I_{\text{as}} = \dfrac{V_s}{R_a'} \]

Where: \(I_{\text{as}}\) = starting armature current (A), \(V_s\) = supply voltage (V), \(R_a'\) = equivalent armature circuit resistance (\(\Omega\))

8.2 With Rheostat

Starting with rheostat inserts variable external resistance into the armature circuit to limit starting current.

Limits initial starting current to safe allowable values
Resistance is cut out in steps as motor accelerates
Starter handle held in running position by hold-fast magnet

Armature Starting Current With Rheostat

\[ I_{\text{as}} = \dfrac{V_s}{R_a' + R} \]

\[ R = \dfrac{V_s}{I_{\text{as}}} - R_a' \]

Where: \(I_{\text{as}}\) = allowable starting current (A), \(V_s\) = supply voltage (V), \(R_a'\) = equivalent armature resistance (\(\Omega\)), \(R\) = starting rheostat resistance (\(\Omega\))

EE Board Exam April 1985

Example. A \(10\ \text{hp}, 220\ \text{V}\) DC shunt motor has an armature and field resistance of \(0.25\ \Omega\) and \(100\ \Omega\) respectively. The full-load efficiency is \(83\%\). Determine the value of starting resistance in order that the starting current will not exceed \(200\%\) of full-load value.

Solution.

\[ P_{\text{in}} = \dfrac{10(746)}{0.83} = 8987.95\ \text{W} \qquad\qquad I_m = \dfrac{8987.95}{220} = 40.854\ \text{A} \]

\[ I_{\text{sh}} = \dfrac{220}{100} = 2.2\ \text{A} \qquad\qquad I_a = 40.854 - 2.2 = 38.654\ \text{A} \]

\[ I_{\text{as}} = 2(38.654) = 77.308\ \text{A} \]

\[ R = \dfrac{V_s}{I_{\text{as}}} - R_a = \dfrac{220}{77.308} - 0.25 = 2.59\ \Omega \]

Answer: \(2.59\ \Omega\)

9. Speed Regulation

Speed regulation is the percentage change in speed when rated mechanical load is reduced to zero.

Measures speed stability of motor under varying loads
Shunt motors have small positive speed regulation
Series motors have extremely high speed regulation

Speed Regulation Formula

\[ \%\text{NR} = \dfrac{N_{\text{NL}} - N_{\text{FL}}}{N_{\text{FL}}} \times 100\% \]

Where: \(\%\text{NR}\) = percentage speed regulation, \(N_{\text{NL}}\) = no-load speed (rpm), \(N_{\text{FL}}\) = full-load speed (rpm)

EE Board Exam October 1990

Example. The nameplate rating of a shunt motor is \(150\ \text{hp}, 600\ \text{V}, 205\ \text{A}, 1700\ \text{rpm}\). The resistance of the shunt field circuit is \(240\ \Omega\) and total armature circuit resistance is \(0.15\ \Omega\). Calculate the speed regulation of the motor.

Solution.

\[ I_{\text{sh}} = \dfrac{600}{240} = 2.5\ \text{A} \qquad\qquad I_{\text{aFL}} = 205 - 2.5 = 202.5\ \text{A} \]

\[ P_{\text{outFL}} = 150(746) = 111900\ \text{W} \qquad\qquad P_{\text{inFL}} = 600(205) = 123000\ \text{W} \]

\[ P_{\text{stray}} = 123000 - 111900 - (202.5)^2(0.15) - 600(2.5) = 3449\ \text{W} \]

\[ I_{\text{aNL}}^2(0.15) - 600 I_{\text{aNL}} + 3449 = 0 \implies I_{\text{aNL}} = 5.757\ \text{A} \]

\[ E_{\text{bNL}} = 600 - 5.757(0.15) = 599.136\ \text{V} \qquad\qquad E_{\text{bFL}} = 600 - 202.5(0.15) = 569.625\ \text{V} \]

\[ N_{\text{NL}} = 1700 \left(\dfrac{599.136}{569.625}\right) = 1788.07\ \text{rpm} \]

\[ \%\text{NR} = \dfrac{1788.07 - 1700}{1700} \times 100\% = 5.18\% \]

Answer: \(5.18\%\)

10. Power Losses and Efficiency

Power losses in a DC motor convert electrical energy into heat and mechanical friction losses.

Armature copper loss: \(P_a = I_a^2 R_a\)
Field copper loss: \(P_{\text{sh}} = V_s I_{\text{sh}}\) and \(P_{\text{se}} = I_{\text{se}}^2 R_{\text{se}}\)
Rotational losses: core hysteresis, eddy current, friction, windage

Motor Power and Efficiency Equations

\[ \eta = \dfrac{P_{\text{output}}}{P_{\text{input}}} \times 100\% \]

\[ \eta = \dfrac{P_{\text{input}} - P_{\text{losses}}}{P_{\text{input}}} \times 100\% \]

\[ P_{\text{input}} = V_s I_m \]

Where: \(\eta\) = overall motor efficiency, \(P_{\text{output}}\) = mechanical shaft power output (W), \(P_{\text{input}}\) = electrical power drawn (W), \(P_{\text{losses}}\) = total power losses (W), \(V_s\) = supply voltage (V), \(I_m\) = line current (A)

EE Board Exam October 1998

Example. The shaft power of a shunt motor is \(7.8\ \text{hp}\). It draws \(50\ \text{A}\) from a \(120\ \text{V}\) line. The field winding draws \(1.2\ \text{A}\). What is the efficiency of the motor?

Solution.

\[ P_{\text{out}} = 7.8(746) = 5818.8\ \text{W} \]

\[ P_{\text{in}} = 120(50) = 6000\ \text{W} \]

\[ \eta = \dfrac{5818.8}{6000} \times 100\% = 96.98\% \]

Answer: \(96.98\%\)

11. Rotational or Constant Losses

Rotational losses encompass magnetic core hysteresis, eddy currents, brush friction, bearing friction, and windage losses.

Remain essentially constant at constant operating speed
Evaluated by running motor without mechanical load
Subtracted from developed power to find net shaft power

Rotational Losses Equation

\[ P_k = V_{\text{sNL}} I_{\text{aNL}} - I_{\text{aNL}}^2 R_a' \]

Where: \(P_k\) = rotational losses (W), \(V_{\text{sNL}}\) = no-load supply voltage (V), \(I_{\text{aNL}}\) = no-load armature current (A), \(R_a'\) = armature resistance (\(\Omega\))

Example. A \(100\ \text{V}\) shunt motor is developing \(6\ \text{hp}\) while operating at an overall efficiency of \(86\%\). The armature and shunt field resistances are \(0.06\ \Omega\) and \(50\ \Omega\) respectively. Determine stray power losses.

Solution.

\[ P_{\text{out}} = 6(746) = 4476\ \text{W} \qquad\qquad P_{\text{in}} = \dfrac{4476}{0.86} = 5204.65\ \text{W} \]

\[ I_m = \dfrac{5204.65}{100} = 52.04\ \text{A} \qquad\qquad I_{\text{sh}} = \dfrac{100}{50} = 2\ \text{A} \]

\[ I_a = 52.04 - 2 = 50.04\ \text{A} \]

\[ P_a = (50.04)^2(0.06) = 150.24\ \text{W} \qquad\qquad P_{\text{sh}} = 100(2) = 200\ \text{W} \]

\[ P_{\text{stray}} = 5204.65 - 4476 - 150.24 - 200 = 378.41\ \text{W} \]

Answer: \(378.41\ \text{W}\)

12. Pony Brake Test of a Motor

Pony brake test is a mechanical friction absorption test used to measure output torque and brake horsepower.

Friction brake band clamped around rotating motor pulley
Scale balance measures net force at known arm radius
Provides direct measurement of mechanical output power

Pony Brake Test Formulas

\[ T = (\text{scale reading} - \text{dead weight}) \times \text{length of arm} \]

\[ \text{Hp} = \dfrac{2 \pi N T}{k} \]

Where: \(T\) = net output torque (\(\text{lb}\cdot\text{ft}\) or \(\text{N}\cdot\text{m}\)), \(N\) = shaft speed (rpm), \(\text{Hp}\) = brake horsepower output (hp), \(k\) = conversion constant (\(33,000\) for \(\text{lb}\cdot\text{ft}\); \(44,760\) for \(\text{N}\cdot\text{m}\))

Example. A pony brake test on a motor running at \(910\ \text{rpm}\) gives a scale reading of \(27.2\ \text{lb}\) with a dead weight of \(2.3\ \text{lb}\). The brake arm length is \(2\ \text{ft}\). If the motor draws \(34\ \text{A}\) from a \(220\ \text{V}\) line, calculate the efficiency.

Solution.

\[ T = (27.2 - 2.3) \times 2 = 49.8\ \text{lb}\cdot\text{ft} \]

\[ \text{Hp}_{\text{out}} = \dfrac{2 \pi (910)(49.8)}{33000} = 8.628\ \text{hp} \qquad\qquad \text{Hp}_{\text{in}} = \dfrac{220(34)}{746} = 10.026\ \text{hp} \]

\[ \eta = \dfrac{8.628}{10.026} \times 100\% = 86.056\% \]

Answer: \(86.056\%\)

13. Formula Summary

Formula Name	Formula	Variables
DC Motor Speed Equations	\(N = \dfrac{60 a E_b}{P Z \Phi}\) \(N = k \dfrac{E_b}{\Phi}\)	\(N\) = speed (rpm), \(E_b\) = back emf (V), \(a\) = parallel current paths, \(P\) = poles, \(Z\) = conductors, \(\Phi\) = flux (Wb), \(k\) = motor constant
Armature Torque Formulas	\(T = \dfrac{P Z \Phi I_a}{2 \pi a}\) \(T = 9.55 \left(\dfrac{P_d}{N}\right)\) \(P_d = E_b I_a\)	\(T\) = torque (\(\text{N}\cdot\text{m}\)), \(P_d\) = developed power (W), \(N\) = speed (rpm), \(E_b\) = back emf (V), \(I_a\) = armature current (A), \(P\) = poles, \(Z\) = conductors, \(\Phi\) = flux (Wb), \(a\) = parallel paths
Mechanical Power Output Formulas	\(\text{Hp} = \dfrac{2 \pi N T}{33000}\) \(\text{Hp} = \dfrac{2 \pi N T}{44760}\)	\(\text{Hp}\) = output power (hp), \(N\) = speed (rpm), \(T\) = torque (\(\text{lb}\cdot\text{ft}\) for 33,000; \(\text{N}\cdot\text{m}\) for 44,760)
Shunt Motor Formulas	\(I_{\text{sh}} = \dfrac{V_s}{R_{\text{sh}}}\) \(I_m = I_a + I_{\text{sh}}\) \(E_b = V_s - I_a R_a'\)	\(I_{\text{sh}}\) = field current (A), \(V_s\) = supply voltage (V), \(R_{\text{sh}}\) = field resistance (\(\Omega\)), \(I_m\) = line current (A), \(I_a\) = armature current (A), \(E_b\) = back emf (V), \(R_a'\) = armature resistance (\(\Omega\))
Series Motor Formulas	\(I_a = I_{\text{se}} = I_m\) \(E_b = V_s - I_a(R_a' + R_{\text{se}})\)	\(I_a\) = armature current (A), \(I_{\text{se}}\) = series current (A), \(I_m\) = line current (A), \(E_b\) = back emf (V), \(V_s\) = voltage (V), \(R_a'\) = armature resistance (\(\Omega\)), \(R_{\text{se}}\) = series resistance (\(\Omega\))
Long-Shunt Compound Motor Formulas	\(I_{\text{se}} = I_a\) \(I_m = I_a + I_{\text{sh}}\) \(I_{\text{sh}} = \dfrac{V_s}{R_{\text{sh}}}\) \(E_b = V_s - I_a(R_a' + R_{\text{se}})\)	\(I_{\text{se}}\) = series current (A), \(I_a\) = armature current (A), \(I_m\) = line current (A), \(I_{\text{sh}}\) = shunt current (A), \(V_s\) = voltage (V), \(R_{\text{sh}}\) = shunt resistance (\(\Omega\)), \(E_b\) = back emf (V), \(R_a'\) = armature resistance (\(\Omega\)), \(R_{\text{se}}\) = series resistance (\(\Omega\))
Short-Shunt Compound Motor Formulas	\(I_{\text{se}} = I_m\) \(I_m = I_a + I_{\text{sh}}\) \(I_{\text{sh}} = \dfrac{V_s - I_{\text{se}} R_{\text{se}}}{R_{\text{sh}}}\) \(E_b = V_s - I_{\text{se}} R_{\text{se}} - I_a R_a'\)	\(I_{\text{se}}\) = series current (A), \(I_m\) = line current (A), \(I_a\) = armature current (A), \(I_{\text{sh}}\) = shunt current (A), \(V_s\) = voltage (V), \(R_{\text{se}}\) = series resistance (\(\Omega\)), \(R_{\text{sh}}\) = shunt resistance (\(\Omega\)), \(E_b\) = back emf (V), \(R_a'\) = armature resistance (\(\Omega\))
Armature Starting Current Without Rheostat	\(I_{\text{as}} = \dfrac{V_s}{R_a'}\)	\(I_{\text{as}}\) = starting current (A), \(V_s\) = voltage (V), \(R_a'\) = armature resistance (\(\Omega\))
Armature Starting Current With Rheostat	\(I_{\text{as}} = \dfrac{V_s}{R_a' + R}\) \(R = \dfrac{V_s}{I_{\text{as}}} - R_a'\)	\(I_{\text{as}}\) = starting current (A), \(V_s\) = voltage (V), \(R_a'\) = armature resistance (\(\Omega\)), \(R\) = rheostat resistance (\(\Omega\))
Speed Regulation Formula	\(\%\text{NR} = \dfrac{N_{\text{NL}} - N_{\text{FL}}}{N_{\text{FL}}} \times 100\%\)	\(\%\text{NR}\) = speed regulation, \(N_{\text{NL}}\) = no-load speed (rpm), \(N_{\text{FL}}\) = full-load speed (rpm)
Motor Power and Efficiency Equations	\(\eta = \dfrac{P_{\text{output}}}{P_{\text{input}}} \times 100\%\) \(\eta = \dfrac{P_{\text{input}} - P_{\text{losses}}}{P_{\text{input}}} \times 100\%\) \(P_{\text{input}} = V_s I_m\)	\(\eta\) = efficiency, \(P_{\text{output}}\) = shaft power (W), \(P_{\text{input}}\) = power drawn (W), \(P_{\text{losses}}\) = total losses (W), \(V_s\) = voltage (V), \(I_m\) = current (A)
Rotational Losses Equation	\(P_k = V_{\text{sNL}} I_{\text{aNL}} - I_{\text{aNL}}^2 R_a'\)	\(P_k\) = rotational losses (W), \(V_{\text{sNL}}\) = no-load voltage (V), \(I_{\text{aNL}}\) = no-load current (A), \(R_a'\) = armature resistance (\(\Omega\))
Pony Brake Test Formulas	\(T = (\text{scale reading} - \text{dead weight}) \times \text{length of arm}\) \(\text{Hp} = \dfrac{2 \pi N T}{k}\)	\(T\) = net torque (\(\text{lb}\cdot\text{ft}\) or \(\text{N}\cdot\text{m}\)), \(N\) = speed (rpm), \(\text{Hp}\) = brake output power (hp), \(k\) = constant (\(33,000\) or \(44,760\))