181. In the figure below, the current of 1 ampere flows through the resistance of—
     
     
     

1. 1. 4 Ω

   
   
   

   2. 20 Ω

   
   
   

   3. 30 Ω

   
   
   

   4. 12 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   In order to know the branch element in which current as flowing 1 amp. first of all we will calculate the voltage drop across the each subgroup. The given circuit is.
   

   VAB = 120 ×	12	= 72 V
   	20

   
   

   VBC =	120 × 2	= 12 V
   	20

   
   

   VCD =	120 × 6	= 36 V
   	20

   
   

   I =	120	= 6 amp
   	20

   
   since voltage drop across BC is 12V.
   

   so current flowing in 12 Ω resistor is =	12	= 1 amp
   	12

   
   Thus, we can say 1 amp current flows through the 12 Ω resistance.

   
   

   Correct Option: D

   In order to know the branch element in which current as flowing 1 amp. first of all we will calculate the voltage drop across the each subgroup. The given circuit is.
   

   VAB = 120 ×	12	= 72 V
   	20

   
   

   VBC =	120 × 2	= 12 V
   	20

   
   

   VCD =	120 × 6	= 36 V
   	20

   
   

   I =	120	= 6 amp
   	20

   
   since voltage drop across BC is 12V.
   

   so current flowing in 12 Ω resistor is =	12	= 1 amp
   	12

   
   Thus, we can say 1 amp current flows through the 12 Ω resistance.

   
   

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182. The equivalent inductance measured between the terminal 1 and 2 for the circuit in figure—
     
     
     

1. 1. L₁ + L₂ + M

   
   
   

   2. L₁ + L₂ – M

   
   
   

   3. L₁ + L₂ + 2M

   
   
   

   4. L_{1 + L2 – 2M}

   
   
   

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1. View Hint View Answer Discuss in Forum

   L_eq = L₁ + L₂ – 2M

   Correct Option: D

   L_eq = L₁ + L₂ – 2M

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183. Impedance Z as shown in figure—
     
     
     

1. 1. J 29 Ω

   
   
   

   2. J 9 Ω

   
   
   

   3. J 19 Ω

   
   
   

   4. J 39 Ω
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   L _eq = L₁ + L₂ + L₃ + 2M₁ – 2M₂
   = J5 + J2 + J2 + 2 × J10 – 2J10
   = J9 + J20 – J20
   = J9 Ω
   

   
   

   Correct Option: B

   L _eq = L₁ + L₂ + L₃ + 2M₁ – 2M₂
   = J5 + J2 + J2 + 2 × J10 – 2J10
   = J9 + J20 – J20
   = J9 Ω
   

   
   

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184. The effective inductance of the circuit across the terminal AB in the figure shown below, is—
     
     
     
     

1. 1. 9 H

   
   
   

   2. 21 H

   
   
   

   3. 11 H

   
   
   

   4. 6 H

   
   
   

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1. View Hint View Answer Discuss in Forum

   L_eq = L₁ + L₂ + L₃ – 2M₁ + 2M₂ – 2M₃
   = 4 + 5 + 6 – 2 × 1 + 2 × 2 – 2 × 3
   = 15 – 2 + 4 – 6
   = 11 H
   

   Correct Option: C

   L_eq = L₁ + L₂ + L₃ – 2M₁ + 2M₂ – 2M₃
   = 4 + 5 + 6 – 2 × 1 + 2 × 2 – 2 × 3
   = 15 – 2 + 4 – 6
   = 11 H
   

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185. The energy stored in the magnetic field at a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current of 10A, is—

1. 1. 0.015 joule

   
   
   

   2. 0.15 joule

   
   
   

   3. 0.5 joule

   
   
   

   4. 1.15 joule

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know
   

   Energy =	µ. N2 A
   	I

   
   

   =	4π × 10–7 × (1000)2 × π × (3 / 2 × 10–2) 2
   	30 × 10–2

   
   = 0.15 joule.

   Correct Option: B

   We know
   

   Energy =	µ. N2 A
   	I

   
   

   =	4π × 10–7 × (1000)2 × π × (3 / 2 × 10–2) 2
   	30 × 10–2

   
   = 0.15 joule.

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