Question
Given the motor and operating conditions described in problem 15.18 , will the armature current increase, decrease, or remain the same, and why?
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18. This includes understanding whether the motor is a DC motor, AC motor, or another type, as well as the specific conditions such as load, speed, and voltage. Show more…
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Explain why the armature current of a shunt motor decreases as the motor accelerates.
In a shunt motor, the permanent magnet is replaced by an electromagnet activated by a field coil that shunts the armature. The shunt motor shown in Fig. 33-4 has an armature resistance of $0.050$ and is connected to a $120 \mathrm{~V}$ line. $(a)$ What is the armature current at the starting instant, i.e., before the armature develops any back emf? (b) What starting rheostat resistance $R$, in series with the armature, will limit the starting current to $60 \mathrm{~A}$ ? $(c)$ With no starting resistance, what back emf is generated when the armature current is $20 \mathrm{~A}$ ? $(d)$ If this machine were running as a generator, what would be the total induced emf developed by the armature when the armature is delivering $20 \mathrm{~A}$ at $120 \mathrm{~V}$ to the shunt field and external circuit? (a) Armature current $=\frac{\text { Impressed voltage }}{\text { Armature resistance }}=\frac{120 \mathrm{~V}}{0.050 \Omega}=2.4 \mathrm{kA}$ (b) Armature current $=\frac{\text { Impressed voltage }}{0.050 \Omega+R} \quad$ or $\quad 60 \mathrm{~A}=\frac{120 \mathrm{~V}}{0.050 \Omega+R}$ from which $R=2.0 \Omega$. (c) Back emf $\begin{aligned}=&(\text { Impressed voltage })-(\text { Voltage drop in armature resistance }) \\ &=120 \mathrm{~V}-(20 \mathrm{~A})(0.050 \Omega)=119 \mathrm{~V}=0.12 \mathrm{kV} \end{aligned}$ (d) $\begin{aligned} \text { Induced emf } &=(\text { Terminal voltage })+(\text { Voltage drop in armature resistance) }\\ &=120 \mathrm{~V}+(20 \mathrm{~A})(0.050 \Omega)=121 \mathrm{~V}=0.12 \mathrm{kV} \end{aligned}$
The shunt motor shown in Fig. 33-5 has an armature resistance of $0.25 \Omega$ and a field resistance of $150 \Omega .$ It is connected across $120-\mathrm{V}$ mains and is generating a back emf of $115 \mathrm{~V}$. Compute: (a) the armature current $I_{a}$, the field current $I_{f}$, and the total current $I_{t}$ taken by the motor; (b) the total power taken by the motor; $(c)$ the power lost in heat in the armature and field circuits; $(d)$ the electrical efficiency of this machine (when only heat losses in the armature and field are considered). $\begin{aligned} \text { (a) } I_{a} &=\frac{(\text { Impressed voltage })-(\text { Back emf })}{\text { Armature resistance }}=\frac{(120-115)}{0.25 \Omega}=20 \mathrm{~A} \\ I_{f} &=\frac{\text { Impressed voltage }}{\text { Field resistance }}=\frac{120 \mathrm{~V}}{150 \Omega}=0.80 \mathrm{~A} \\ I_{t} &=I_{a}+I_{f}=20.80 \mathrm{~A}=21 \mathrm{~A} \end{aligned}$ (b) Power input $=(120 \mathrm{~V})(20.80 \mathrm{~A})=2.5 \mathrm{~kW}$ (c) $I_{a}^{2} r_{a}$ loss in armature $=(20 \mathrm{~A})^{2}(0.25 \Omega)=0.10 \mathrm{~kW}$ $I_{f}^{2} r_{f}$ loss in field $=(0.80 \mathrm{~A})^{2}(150 \Omega=96 \mathrm{~W}$ (d) Power output $=($ Power input $)-($ Power losses $)=2496-(100+96)=2.3 \mathrm{~kW}$ Alternatively. Power output $=($ Armature current $)($ Back $\mathrm{emf})=(20 \mathrm{~A})(115 \mathrm{~V})=2.3 \mathrm{~kW}$ Then $\quad$ Efficiency $=\frac{\text { Power output }}{\text { Power input }}=\frac{2300 \mathrm{~W}}{2496 \mathrm{~W}}=0.921=92 \%$
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