36. A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be​​

1. 1. 4.0 A

   
   
   

   2. 8.0 A ​

   
   
   

   3. 	20	A
      	√13

   
   
   

   4. 2.0 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   If ω = 50 × 2π then ωL = 20Ω ​
   If ω′ = 100 × 2π then ω′L = 40Ω ​
   Current flowing in the coil is
   

   I =	200	=	200	=	200
   	Z		√R² + (ω'L)²		√(30)² + (40)²

   
   I = 4A.

   Correct Option: A

   If ω = 50 × 2π then ωL = 20Ω ​
   If ω′ = 100 × 2π then ω′L = 40Ω ​
   Current flowing in the coil is
   

   I =	200	=	200	=	200
   	Z		√R² + (ω'L)²		√(30)² + (40)²

   
   I = 4A.

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37. An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is​​

1. 1. π/6 ​

   
   
   

   2. π/4 ​

   
   
   

   3. ​π/2 ​

   
   
   

   4. zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   The phase difference φ is given by ​
   

   tan φ =	XL
   	R

   
   

   =	3	= 1 ⇒ φ =	π
   	3		4

   
   

   Correct Option: B

   The phase difference φ is given by ​
   

   tan φ =	XL
   	R

   
   

   =	3	= 1 ⇒ φ =	π
   	3		4

   
   

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38. In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage the current in the circuit is  π/3. If instead, C is removed from the circuit, the phase difference is again π/3. The power factor of the circuit is :​​

1. 1. ​1/2 ​

   
   
   

   2. 1/√2

   
   
   

   3. 1

   
   
   

   4. √3 / 2
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   when L is removed from the circuit
   

   	Xc	= tan	π
   	R		3

   
   

   Xc = R tan	π	...(1)
   	3

   
   
   when C is remove from the circuit
   

   	XL	= tan	π
   	R		3

   
   

   Xc = R tan	π	...(2)
   	3

   
   
   net impedence Z = √R² + (X_L - X_C)² = R
   

   power factor cos φ =	R	= 1
   	Z

   
   

   Correct Option: C

   
   when L is removed from the circuit
   

   	Xc	= tan	π
   	R		3

   
   

   Xc = R tan	π	...(1)
   	3

   
   
   when C is remove from the circuit
   

   	XL	= tan	π
   	R		3

   
   

   Xc = R tan	π	...(2)
   	3

   
   
   net impedence Z = √R² + (X_L - X_C)² = R
   

   power factor cos φ =	R	= 1
   	Z

   
   

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39. A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when​​

1. 1. number of turns in the coil is reduced

   
   
   

   2. a capacitance of reactance X_c = X_L is included in the same circuit

   
   
   

   3. an iron rod is inserted in the coil

   
   
   

   4. frequency of the Ac source is decreased

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​By inserting iron rod in the coil, ​L ↑ z ↑ I ↓ so brightness ↓
   

   Correct Option: C

   ​By inserting iron rod in the coil, ​L ↑ z ↑ I ↓ so brightness ↓
   

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40. A series R-C circuit is connected to an alternating voltage source. Consider two situations:​​
    (A)​When capacitor is air filled. ​
    (B)​When capacitor is mica filled. ​
    Current through resistor is i and voltage across capacitor is V then : ​

1. 1. V_a > V_b

   
   
   

   2. ​i_a > i_b

   
   
   

   3. V_a = V_b

   
   
   

   4. ​V_a < V_b

   
   
   

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1. View Hint View Answer Discuss in Forum

   For series R – C circuit, capacitive reactance,
   Z_c = √R² + (1 / Cω)²
   
   

   current i =	V	=	V
   	Zc		√R² + (1/Cω)²

   
   

   Vc = iXc =	V	×	1
   	√ R² + (1 / Cω)²		(1/Cω)

   
   V_c = V / √(R Cω)² + 1
   If we fill a di-electric material like mica instead of air then  capacitance C↑ ⇒ Vc↓ ​So, V_a > V_b
   

   Correct Option: A

   For series R – C circuit, capacitive reactance,
   Z_c = √R² + (1 / Cω)²
   
   

   current i =	V	=	V
   	Zc		√R² + (1/Cω)²

   
   

   Vc = iXc =	V	×	1
   	√ R² + (1 / Cω)²		(1/Cω)

   
   V_c = V / √(R Cω)² + 1
   If we fill a di-electric material like mica instead of air then  capacitance C↑ ⇒ Vc↓ ​So, V_a > V_b
   

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