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    GATE (91) PART B, PROBLEM-13 

Soln:- The above problem has really puzzled me, 
Equivalent ckt consists of a 50 source opposing a 10-60 source and series resistance of (6+4). 
This gives i = (10-60 - 50 )/(6+4) = -j0.866 amps (anticlockwise sense) 

Using voltage conjugate VI*, 

Power 50 source………….( 50 )( -j0.866 )* (absorbed) 
= j(4.33) 

Power 10-60 source……….(10-60 )( -j0.866 )* (generated) 
= 7.5 + j(4.33) 

It indicates 7.5 W is dissipated in the (6+4). 
Which gives, P6 = 4.5 W 
P4 = 3.0 W 

But the actual answer is different, 

Power dissipated in the resistors = 0 (since real part of current is zero) 

Power delivered by 50 source = 2.5W 
Power delivered by 10-60 source = 10W 

GATE (95), SECTION (B), Q-14, 

Electromagnet problem……. 

• How would u calculate  ? 
max = 230/(4.44 f N) 
rms = max/2 

Or 
230/Rg (as provided by Handa’s solution) 

The answer does not come if the former is applied. 

GATE (99) SEC-A, 1.9 

What does “3db bandwidth of 5hz” mean in this question? 

GATE (2000) SEC-B 4(b) 

My answer: - B1/ (82) 
Gate forum answer: - B1/8 
Handa’s solution: - B1/ (22) 

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balu Senior Member Joined: 20Feb2007 Online Status: Offline Posts: 236	Topic: network problems, please share your views!! Posted: 22Feb2007 at 4:08pm
	GATE (91) PART B, PROBLEM-13 Soln:- The above problem has really puzzled me, Equivalent ckt consists of a 50 source opposing a 10-60 source and series resistance of (6+4). This gives i = (10-60 - 50 )/(6+4) = -j0.866 amps (anticlockwise sense) Using voltage conjugate VI, Power 50 source………….( 50 )( -j0.866 ) (absorbed) = j(4.33) Power 10-60 source……….(10-60 )( -j0.866 )* (generated) = 7.5 + j(4.33) It indicates 7.5 W is dissipated in the (6+4). Which gives, P6 = 4.5 W P4 = 3.0 W But the actual answer is different, Power dissipated in the resistors = 0 (since real part of current is zero) Power delivered by 50 source = 2.5W Power delivered by 10-60 source = 10W GATE (95), SECTION (B), Q-14, Electromagnet problem……. • How would u calculate  ? max = 230/(4.44 f N) rms = max/2 Or 230/Rg (as provided by Handa’s solution) The answer does not come if the former is applied. GATE (99) SEC-A, 1.9 What does “3db bandwidth of 5hz” mean in this question? GATE (2000) SEC-B 4(b) My answer: - B1/ (82) Gate forum answer: - B1/8 Handa’s solution: - B1/ (22) Post Resume: Click here to Upload your Resume & Apply for Jobs

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