A transformer has core loss = 500 W and full-load copper loss = 500 W.
1. Maximum efficiency occurs at full load condition.
2. At half load, copper loss becomes 125 W.
3. Efficiency at half load is lower than full load efficiency.
4. Condition for maximum efficiency is that copper loss equals core loss.
Which of the above statements is/are correct?
Answer & step-by-step solution
At x p.u. load, transformer output is proportional to x while copper loss is proportional to x² and iron loss remains approximately constant. The given losses make the maximum-efficiency load x=sqrt(500/500)=1 p.u. Every statement follows numerically, so B is correct. Key relation: Pcu(x)=x²Pcu,FL. Here Pcu(0.5)=0.25×500=125 W. Maximum η: x²Pcu,FL=Pi → x=1.
Exam focus: If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result. Common trap: Scaling copper loss linearly with load instead of with load squared.
Statement-wise Verification
- Statement 1 - Correct.
Maximum efficiency occurs at full load condition.
Maximum efficiency occurs when Pcu=Pi; this equality is given at full load. - Statement 2 - Correct.
At half load, copper loss becomes 125 W.
Copper loss scales with current/load squared, so at 0.5 load it is 0.5²×500=125 W. - Statement 3 - Correct.
Efficiency at half load is lower than full load efficiency.
Full load is the maximum-efficiency operating point in this case, so moving to half load gives lower efficiency (for the same power factor assumption). - Statement 4 - Correct.
Condition for maximum efficiency is that copper loss equals core loss.
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Core Concept
At x p.u. load, transformer output is proportional to x while copper loss is proportional to x² and iron loss remains approximately constant. The efficiency curve therefore has a peak when the two loss components are equal.
Formula / Key Relationship
Pcu(x)=x²Pcu,FL. Here Pcu(0.5)=0.25×500=125 W. Maximum η: x²Pcu,FL=Pi → x=1.
Step-by-Step Check
The given losses make the maximum-efficiency load x=sqrt(500/500)=1 p.u. Every statement follows numerically, so B is correct.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 4. Option D is incorrect because it omits correct statement(s) 3.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Common Mistake
Scaling copper loss linearly with load instead of with load squared.
Quick Revision
Load for Maximum Efficiency is linked with Transformer Efficiency, Transformers and Electrical Machines. If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
एक ट्रांसफार्मर की कोर हानि = 500 W और पूर्ण-लोड कॉपर हानि = 500 W है।
1. अधिकतम दक्षता पूर्ण लोड पर होती है।
2. आधे लोड पर कॉपर हानि 125 W हो जाती है।
3. आधे लोड पर दक्षता पूर्ण लोड दक्षता से कम है।
4. अधिकतम दक्षता की शर्त है कि कॉपर हानि कोर हानि के बराबर हो।
उपरोक्त में से कौन-सा/से कथन सही है/हैं?
उत्तर और चरणबद्ध हल
x p.u. लोड पर आउटपुट x के, कॉपर हानि x² के समानुपाती और कोर हानि लगभग स्थिर है। दी गई हानियों से अधिकतम-दक्षता लोड x=√(500/500)=1 p.u.। Pcu(0.5)=0.25×500=125 W। सभी कथन संख्यात्मक रूप से सही हैं, अतः B।
परीक्षा फोकस: कोर हानि = पूर्ण-लोड कॉपर हानि हो तो अधिकतम दक्षता पूर्ण लोड पर। सामान्य गलती: कॉपर हानि को लोड के साथ रैखिक मानना।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
Maximum efficiency occurs at full load condition.
Maximum efficiency occurs when Pcu=Pi; this equality is given at full load. - Statement 2 - Correct.
At half load, copper loss becomes 125 W.
Copper loss scales with current/load squared, so at 0.5 load it is 0.5²×500=125 W. - Statement 3 - Correct.
Efficiency at half load is lower than full load efficiency.
Full load is the maximum-efficiency operating point in this case, so moving to half load gives lower efficiency (for the same power factor assumption). - Statement 4 - Correct.
Condition for maximum efficiency is that copper loss equals core loss.
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Core Concept
At x p.u. load, transformer output is proportional to x while copper loss is proportional to x² and iron loss remains approximately constant. The efficiency curve therefore has a peak when the two loss components are equal.
Formula / Key Relationship
Pcu(x)=x²Pcu,FL. Here Pcu(0.5)=0.25×500=125 W. Maximum η: x²Pcu,FL=Pi → x=1.
Step-by-Step Check
The given losses make the maximum-efficiency load x=sqrt(500/500)=1 p.u. Every statement follows numerically, so B is correct.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 4. Option D is incorrect because it omits correct statement(s) 3.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Common Mistake
Scaling copper loss linearly with load instead of with load squared.
Quick Revision
Load for Maximum Efficiency is linked with Transformer Efficiency, Transformers and Electrical Machines. If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
એક ટ્રાન્સફોર્મરનો કોર લોસ = 500 W અને સંપૂર્ણ-લોડ કોપર લોસ = 500 W છે.
1. મહત્તમ કાર્યક્ષમતા સંપૂર્ણ લોડ પર મળે છે.
2. અડધા લોડ પર કોપર લોસ 125 W થાય છે.
3. અડધા લોડ પર કાર્યક્ષમતા સંપૂર્ણ લોડ કાર્યક્ષમતા કરતાં ઓછી છે.
4. મહત્તમ કાર્યક્ષમતાની શરત છે કે કોપર લોસ કોર લોસ બરાબર હોય.
ઉપરોક્તમાંથી કયું/કયા વિધાન સાચું/સાચા છે?
જવાબ અને પગલાવાર ઉકેલ
x p.u. લોડ પર આઉટપુટ x ના, કોપર લોસ x² ના સમપ્રમાણમાં અને કોર લોસ લગભગ અચળ છે. આપેલ લોસ પરથી મહત્તમ-કાર્યક્ષમતા લોડ x=√(500/500)=1 p.u.. Pcu(0.5)=0.25×500=125 W. બધાં વિધાનો આંકડાકીય રીતે સાચાં છે, તેથી B.
પરીક્ષા મુદ્દો: કોર લોસ = સંપૂર્ણ-લોડ કોપર લોસ હોય તો મહત્તમ કાર્યક્ષમતા સંપૂર્ણ લોડ પર. સામાન્ય ભૂલ: કોપર લોસને લોડ સાથે રેખીય માનવો.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
Maximum efficiency occurs at full load condition.
Maximum efficiency occurs when Pcu=Pi; this equality is given at full load. - Statement 2 - Correct.
At half load, copper loss becomes 125 W.
Copper loss scales with current/load squared, so at 0.5 load it is 0.5²×500=125 W. - Statement 3 - Correct.
Efficiency at half load is lower than full load efficiency.
Full load is the maximum-efficiency operating point in this case, so moving to half load gives lower efficiency (for the same power factor assumption). - Statement 4 - Correct.
Condition for maximum efficiency is that copper loss equals core loss.
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Core Concept
At x p.u. load, transformer output is proportional to x while copper loss is proportional to x² and iron loss remains approximately constant. The efficiency curve therefore has a peak when the two loss components are equal.
Formula / Key Relationship
Pcu(x)=x²Pcu,FL. Here Pcu(0.5)=0.25×500=125 W. Maximum η: x²Pcu,FL=Pi → x=1.
Step-by-Step Check
The given losses make the maximum-efficiency load x=sqrt(500/500)=1 p.u. Every statement follows numerically, so B is correct.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 4. Option D is incorrect because it omits correct statement(s) 3.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
Common Mistake
Scaling copper loss linearly with load instead of with load squared.
Quick Revision
Load for Maximum Efficiency is linked with Transformer Efficiency, Transformers and Electrical Machines. If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result.
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