31. The primary of a transformer when connected to a dc battery of 10 volt draws a current of 1 mA. The number of turns of the primary and secondary windings are 50 and 100 respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively​​

1. 1. 20 V and 0.5 mA ​

   
   
   

   2. 20 V and 2.0 mA ​

   
   
   

   3. ​10 V and 0.5 mA ​

   
   
   

   4. Zero and therefore no current

   
   
   

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1. View Hint View Answer Discuss in Forum

   A transformer is essentially an AC device. DC source so no mutual induction between coils
   ⇒ E₂ = 0 and I₂ = 0
   

   Correct Option: D

   A transformer is essentially an AC device. DC source so no mutual induction between coils
   ⇒ E₂ = 0 and I₂ = 0
   

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32. ​What is the value of inductance L for which the current is maximum in a series LCR circuit with ​C = 10 µF and ω = 1000s^{– 1} ?​​

1. 1. 1 mH ​

   
   
   

   2. ​cannot be calculated unless R is known

   
   
   

   3. 10 mH ​

   
   
   

   4. 100 mH

   
   
   

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1. View Hint View Answer Discuss in Forum

   Condition for which the current is maximum in a series LCR circuit is,
   

   ω =	1
   	√LC

   
   

   1000 =	1
   	√L(10 × 10 -6)

   
   ⇒  = L = 100 mH

   Correct Option: D

   Condition for which the current is maximum in a series LCR circuit is,
   

   ω =	1
   	√LC

   
   

   1000 =	1
   	√L(10 × 10 -6)

   
   ⇒  = L = 100 mH

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33. ​Power dissipated in an LCR series circuit connected to an a.c source of emf ε is​

1. 1. 	ε²√R² + [Lω - (1/Cω)]²
      	R

   
   
   

   2. 	ε²[R² + Lω - (1/Cω)]²
      	R

   
   
   

   3. 	ε²/R
      	R² + [Lω - (1/Cω)]²

   
   
   

   4. 	ε²/R
      	[R² + Lω - (1/Cω)]²

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Power  dissipated in series LCR;
   

   P = I² R =	ε ²	R
   	(Z)²

   
   

   =	ε ² R
   	[R² + (ωL -(1/ωC))²]

   
   Where Z = √R² + [ωL -(1 / ωC)]²
   is called the impedance of the circuit.

   Correct Option: D

   Power  dissipated in series LCR;
   

   P = I² R =	ε ²	R
   	(Z)²

   
   

   =	ε ² R
   	[R² + (ωL -(1/ωC))²]

   
   Where Z = √R² + [ωL -(1 / ωC)]²
   is called the impedance of the circuit.

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34. A condenser of capacity C is charged to a potential difference of V1. The plates of the  condenser are
    then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V₂ is​

1. 1. [C(V₁² - V₂²)/L]^1/2

   
   
   

   2. [C(V₁ - V₂)²/L]^1/2

   
   
   

   3. [C(V₁² - V₂²)/L]

   
   
   

   4. [CV₁ - V₂)/L]

   
   
   

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1. View Hint View Answer Discuss in Forum

   q = CV₁ cos wt
   

   ⇒ i =	dq	= - ω Cv1 sin ωt
   	dt

   
   

   Also, ω² =	1	and V = V1 cos ωt
   	LC

   
   At t = t₁, V = V₂ and i = -ω CV₁ sinωt₁
   

   ∴ cos ωt1 =	V2	(–ve sign gives direction)
   	V1

   
   
   = {C(V₁²) - V₂²) / L}²
   

   Correct Option: A

   q = CV₁ cos wt
   

   ⇒ i =	dq	= - ω Cv1 sin ωt
   	dt

   
   

   Also, ω² =	1	and V = V1 cos ωt
   	LC

   
   At t = t₁, V = V₂ and i = -ω CV₁ sinωt₁
   

   ∴ cos ωt1 =	V2	(–ve sign gives direction)
   	V1

   
   
   = {C(V₁²) - V₂²) / L}²
   

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35. In the given circuit the reading of voltmeter V₁ and V₂ are 300 volts each. The reading of the voltmeter V₃ and ammeter A are respectively​​
    

1. 1. 150 V, 2.2 A ​

   
   
   

   2. 220 V, 2.2 A ​

   
   
   

   3. ​220 V, 2.0 A ​

   
   
   

   4. 100 V, 2.0 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​As V_L = V_C = 300 V, resonance will take place ​​
   ∴ V_R = 220 V
   Current, I = 220/100 = 2.2A
   ∴ reading of V₃ = 220 V ​
   and reading of A = 2.2 A

   Correct Option: B

   ​As V_L = V_C = 300 V, resonance will take place ​​
   ∴ V_R = 220 V
   Current, I = 220/100 = 2.2A
   ∴ reading of V₃ = 220 V ​
   and reading of A = 2.2 A

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