DC Generator Efficiency MCQ Quiz in తెలుగు - Objective Question with Answer for DC Generator Efficiency - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Apr 19, 2025
Latest DC Generator Efficiency MCQ Objective Questions
Top DC Generator Efficiency MCQ Objective Questions
DC Generator Efficiency Question 1:
A 125 V, 12.5 kW shunt generator is driven by a 20hp motor to generate rated output. The armature circuit resistance is 0.1 Ω, and the field circuit resistance is 62.5 Ω. The rated variable electric loss is 1040 W. At rated power output, the armature current for maximum efficiency is__ (in A)
Answer (Detailed Solution Below) 117 - 118
DC Generator Efficiency Question 1 Detailed Solution
Shunt field loss,\(\frac{{V_t^2}}{{{R_{sh}}}} = \frac{{{{125}^2}}}{{62.5}} = 250\;W\)
Load current,\({I_L} = \frac{{{P_{out}}}}{{{V_t}}} = \frac{{12500}}{{125}} = 100\;A\)
Shunt field current,\({I_{sh}} = \frac{{{V_t}}}{{{R_{sh}}}} = \frac{{125}}{{62.5}} = 2\;A\)
Armature current,\({I_a} = {I_L} + {I_{sh}} = 102\;A\)
Generated emf,\({E_g} = {V_t} + {I_a}{R_a} = 125 + \left( {102 \times 0.1} \right) = 135.2\;V\)
Developed power,\({P_d} = {E_g}{I_a} = 135.2 \times 102 = 13790.4\;W\)
Rotational losses,\({P_r} = {P_{in}} - {P_d} = \left( {20 \times 746} \right) - \left( {13790.4} \right) = 1129.6\;W\)
The constant loss is,\({P_{const}} = {P_r} + {V_{sh}}{I_{sh}} = 1129.6 + 250 = 1379.6\;W\)
For maximum efficiency, armature current
\({I_a} = \sqrt {\frac{{{P_{const}}}}{{{R_a}}}} = \sqrt {\frac{{1379.6}}{{0.1}}} = 117.4\;A\)
DC Generator Efficiency Question 2:
A shunt generator delivers 160 A at terminal voltage of 220V. The armature resistance and shunt field resistance are 0.06 and 90 Ω respectively. The iron and friction losses are 700 W. What are mechanical, commercial and electrical efficiencies for generator respectively?
Answer (Detailed Solution Below)
98.16%, 92.58%, 94.32%
DC Generator Efficiency Question 2 Detailed Solution
Current in shunt field, \({I_{sh}} = \frac{{220}}{{90}}\)
= 2.44 A
Armature emf, E = V + Iara
= 220 + (160 + 2.44) × 0.06
= 229.75 V
Armature copper loss \( = I_a^2{r_a}\)
\( = {\left( {162.44} \right)^2} \times 0.06\)
= 1583 W
Shunt Cu loss = VIsh
= 220 × 2.44
= 536.8 W
Total cu loss = 2119.8 W
Stray losses = 700 W
Total losses = 2819.8 W
o/P=VI=220×160
= 35200 W
I/P Provided by prime mover
= 35200 + 2819.8
= 38019.8 W
= Generator i/P
Electrical power produced in armature
= EgIa = 38019.8 – 700
= 37319.8 W
Mechanical Efficiency \( = \frac{{{\varepsilon _g}{I_a}}}{{Prime\ mover\frac{I}{P}}} = \frac{{37319.8}}{{38019.8}}\)
= 0.9816
= 98.16%
Electrical efficiency \( = \frac{{{V_I}}}{{{E_g}{I_a}}} = \frac{{35200}}{{37319.8}}\)
= 94.32%
Commercial efficiency \( = \frac{{35200}}{{38019.8}}\)
= 92.58%