181. A two terminal black box contains one of the R.L.C. elements. The black box is connected to a 220 volts A.C. supply. The current through the source is I. When a capacitance of 0.1F is inserted in the series between the source and the box, the current through the source is 2I. The element is
     

1. 1. a resistance

   
   
   

   2. an inductance

   
   
   

   3. a capacitance of 0.5F

   
   
   

   4. it is not possible to determine the element.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Initially , I =	V
   	R + jX

   
   When capacitance inserted,
   

   2I =	V
   	R + jX - jωC

   
   

   ⇒	2	+	6
   	R + jX		R + jX - (j / ωC)

   
   

   ⇒ 2R + 2jX -	2j	= R + jX
   	ωC

   
   

   ⇒ R + jX =	2j
   	ωC

   
   

   X =	2
   	ωC

   
   Here ‘+’ ve sign element is inductance
   

   Correct Option: B

   Initially , I =	V
   	R + jX

   
   When capacitance inserted,
   

   2I =	V
   	R + jX - jωC

   
   

   ⇒	2	+	6
   	R + jX		R + jX - (j / ωC)

   
   

   ⇒ 2R + 2jX -	2j	= R + jX
   	ωC

   
   

   ⇒ R + jX =	2j
   	ωC

   
   

   X =	2
   	ωC

   
   Here ‘+’ ve sign element is inductance
   

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182. In the transformer shown in the given figure, the inductance measured across the terminal 1 and 2 was 4 H with open terminals 3 and 4. It was 3 H when the terminal 3 and 4 are were short circuited. The coefficient of coupling would be
     
     

1. 1. 1

   
   
   

   2. 0.707

   
   
   

   3. 0.5

   
   
   

   4. indeterminate due to insufficient data

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA
   

   Correct Option: D

   NA
   

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183. The value of the capacitance C in the given ac circuit to make it a constant resistance circuit OR for the supply current to be independent of its frequency is
     
     

1. 1. 	1	F
      	16

      
      

   
   
   

   2. 	1	F
      	12

      
      

   
   
   

   3. 	1	F
      	8

      
      

   
   
   

   4. 	1	F
      	4

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   
   For constant resistance, equating imaginary term to zero,
   

   16 +

   1

   ω -

   1

   = 32

   ω -

   1

   C

   ωC

   ωC

   
   

   ∴ C =	1	F
   	16

   Correct Option: A

   
   
   For constant resistance, equating imaginary term to zero,
   

   16 +

   1

   ω -

   1

   = 32

   ω -

   1

   C

   ωC

   ωC

   
   

   ∴ C =	1	F
   	16

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184. A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled, then the voltage across each resistor is
     

1. 1. halved

   
   
   

   2. doubled

   
   
   

   3. increased by four times

   
   
   

   4. not changed

   
   
   

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1. View Hint View Answer Discuss in Forum

   First Method :
   

   	V	ratio is a constant R. If R is doubled current will become half.
   	I

   
   

   V = 2R	I	= RI
   	2

   
   Second Method :
   

   VR1 =	V	× R1
   	R1 + R2

   
   

   VR2 =	V	× R2
   	R1 + R2

   
   If R₁ and R₂ are doubled, then
   

   V'R1 =	V × 2R1	= VR1
   	2(R1 + R2)

   
   

   V'R2 =	V × 2R1	= VR2
   	2(R1 + R2)

   
   

   Correct Option: D

   First Method :
   

   	V	ratio is a constant R. If R is doubled current will become half.
   	I

   
   

   V = 2R	I	= RI
   	2

   
   Second Method :
   

   VR1 =	V	× R1
   	R1 + R2

   
   

   VR2 =	V	× R2
   	R1 + R2

   
   If R₁ and R₂ are doubled, then
   

   V'R1 =	V × 2R1	= VR1
   	2(R1 + R2)

   
   

   V'R2 =	V × 2R1	= VR2
   	2(R1 + R2)

   
   

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185. Current I₁, I₂ and I₃ meet at a junction (node) in a circuit. All currents are marked as entering the node. If I₁ = – 6 sin (ω t) mA and I₂ = 8 cos (ω t) mA, then I 3 will be
     

1. 1. 10 cos (ωt + 36.87) mA

   
   
   

   2. 14 cos (ωt + 36.87) mA

   
   
   

   3. – 14 sin (ωt + 36.87) mA

   
   
   

   4. 10 cos (ωt + 36.87) mA

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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