A 10 b.h.p. 7.46 kW, 200- V shunt motor has full-load efficiency of 85 % . The armature has a resistance of 0.238 Ω . Calculate the value of the starting resistance necessary to limit the starting current to 1.5 times the full-load current at the moment of first switching on. The shunt current

Question

A 10 b.h.p. 7.46 kW, 200- V shunt motor has full-load efficiency of $85 \% .$ The armature has a resistance of $0.238 \Omega .$ Calculate the value of the starting resistance necessary to limit the starting current to 1.5 times the full-load current at the moment of first switching on. The shunt current may be neglected. Find also the back emf of the motor, when the current has fallen to its full-load value, assuming that the whole of the starting resistance is still in circuit.

Accepted Answer

Full load motor current, [latex] I_{f l}=\frac{7.46 imes 1000}{200 imes 0.85}=43.88 A [/latex]

Starting current, [latex] I_{1}=1.5 imes 43.88=65.83 A [/latex]

[latex] R_{1}=\frac{V}{I_{1}}=\frac{200}{65.83}=3.038 \Omega ; R_{a}=0.238 \Omega [/latex]

[latex] herefore [/latex] Starting resistance [latex] =R_{1}-R_{a}=3.038-0.238=2.8 \Omega [/latex]

[latex] ext { Now, } \quad ext { full-load current, } I_{f l}=43.88 A =I_{a}[/latex]

[latex] ext { Now, } \quad I_{a}=\frac{V-E_{b}}{R_{1}}[/latex]

[latex] herefore \quad E_{b}=V-I_{a} R_{1}=200-(43.88 imes 3.038)=67 V [/latex]

Question 5.12: A 10 b.h.p. 7.46 kW, 200- V shunt motor has full-load effici...

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