Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

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 Step 1

Given circuit:

$$\begin{gathered} \mathrm{Given }: \\ {\mathrm{V }}_{\mathrm{L }}= \mathrm{12 }\mathrm{V } \\ {\mathrm{I }}_{\mathrm{L }}= \left(\mathrm{0 }- \mathrm{200 }\mathrm{mA }\right) \\ \mathrm{For }\mathrm{maintaining }\mathrm{load }\mathrm{voltage }, {\mathrm{V }}_{\mathrm{L }}= \mathrm{12 }\mathrm{V }, \mathrm{we }\mathrm{will }\mathrm{use }\mathrm{zener }\mathrm{diode }\mathrm{of }{\mathrm{V }}_{\mathrm{Z }}= \mathrm{12 }\mathrm{V } \\ \mathbf{\therefore }{\mathbf{V }}_{\mathbf{Z }}\mathbf{= }\mathbf{12 }\mathbf{}\mathbf{V} \end{gathered}$$

Step 2

$$\begin{gathered} \mathrm{For }\mathrm{calculating }\mathrm{load }\mathrm{resistance }{\mathrm{R }}_{\mathrm{L }}: \\ {\mathrm{R }}_{\mathrm{L }}= \frac{{\mathrm{V }}_{\mathrm{Z }}}{{\mathrm{I }}_{\mathrm{L }}} \\ \mathrm{Here }, \\ \mathrm{Zener }\mathrm{voltage }{\mathrm{V }}_{\mathrm{Z }}\mathrm{is }\mathrm{12 }\mathrm{V } \\ \mathrm{Load }\mathrm{current }{\mathrm{I }}_{\mathrm{L }}\mathrm{is }\mathrm{200 }\mathrm{mA } \\ \mathrm{Substitute }\mathrm{12 }\mathrm{V }\mathrm{for }{\mathrm{V }}_{\mathrm{Z }}\mathrm{and }\mathrm{200 }\mathrm{mA }\mathrm{for }{\mathrm{I }}_{\mathrm{L }}\mathrm{to }\mathrm{obtain }{\mathrm{R }}_{\mathrm{L }}. \\ {\mathrm{R }}_{\mathrm{L }}= \frac{{\mathrm{V }}_{\mathrm{Z }}}{{\mathrm{I }}_{\mathrm{L }}} \\ {\mathrm{R }}_{\mathrm{L }}= \frac{\mathrm{12 }\mathrm{V }}{\mathrm{200 }\mathrm{mA }} \\ {\mathbf{R }}_{\mathbf{L }}\mathbf{= }\mathbf{60 }\mathbf{\Omega } \end{gathered}$$

Step 3

$$\begin{gathered} \mathrm{Now }, \mathrm{the }\mathrm{source }\mathrm{resistance }{\mathrm{R }}_{\mathrm{S }}; \\ {\mathrm{V }}_{\mathrm{L }}= \frac{{\mathrm{R }}_{\mathrm{L }}{\mathrm{V }}_{\mathrm{i }}}{{\mathrm{R }}_{\mathrm{L }}+ {\mathrm{R }}_{\mathrm{S }}} \\ \mathrm{Here }, \\ \mathrm{Input }\mathrm{voltage }{\mathrm{V }}_{\mathrm{i }}\mathrm{is }\mathrm{16 }\mathrm{V } \\ \mathrm{Load }\mathrm{voltage }{\mathrm{V }}_{\mathrm{L }}\mathrm{is }\mathrm{12 }\mathrm{V } \\ \mathrm{Substitute }\mathrm{16 }\mathrm{V }\mathrm{for }{\mathrm{V }}_{\mathrm{i }}, \mathrm{60 }\mathrm{\Omega }\mathrm{for }{\mathrm{R }}_{\mathrm{L }}\mathrm{and }\mathrm{12 }\mathrm{V }\mathrm{for }{\mathrm{V }}_{\mathrm{L }}\mathrm{to }\mathrm{obtain }{\mathrm{R }}_{\mathrm{S }}. \\ \mathrm{12 }= \frac{\left(\mathrm{60 }\right)\left(\mathrm{16 }\right)}{\mathrm{60 }+ {\mathrm{R }}_{\mathrm{S }}} \\ \mathrm{720 }+ \mathrm{12 }{\mathrm{R }}_{\mathrm{S }}= \mathrm{960 } \\ \mathrm{12 }{\mathrm{R }}_{\mathrm{S }}= \mathrm{240 } \\ {\mathrm{R }}_{\mathrm{S }}= \mathrm{20 }\mathrm{\Omega } \\ \therefore \mathrm{The }\mathrm{value }\mathrm{of }\mathrm{the }\mathrm{source }\mathrm{resistance }{\mathrm{R }}_{\mathrm{S }}\mathrm{is }\mathbf{}\mathbf{20 }\mathbf{\Omega } \end{gathered}$$