61. Five cells are connected in series in a row and then four such rows are connected in parallel to feed the current to a resistive load of 1.25 Ω . Each cell has emf of 1.5 V with internal resistance of 0.2 Ω . The current through the load will be _____ .
    

1. 1. 1 A

   
   
   

   2. 5 A

   
   
   

   3. 15 A

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   Req =	5 × 0.2	= 0.25
   	4

   
   

   I =	5 . E	=	5 × 1.5	= 5 A
   	RL + Req		1.25 + 0.25

   
   

   Correct Option: B

   Req =	5 × 0.2	= 0.25
   	4

   
   

   I =	5 . E	=	5 × 1.5	= 5 A
   	RL + Req		1.25 + 0.25

   
   

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62. When two coupled coils of equal self-inductance are connected in series in one way the net inductance is 12 mH, and when they are connected in the other way, the net inductance is 4 mH. The maxiumum value of net inductance when they are connected in parallel in a suitable way is _________ mH
    

1. 1. 3 mH

   
   
   

   2. 5 mH

   
   
   

   3. 9 mH

   
   
   

   4. 10 mH

   
   
   

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1. View Hint View Answer Discuss in Forum

   As, 2L + 2M = 12
   2L – 2M = 4
   Then, L = 4
   ⇒ M = 2
   To get maximum value in parallel connection,
   

   Lmax =	(L + M)(L + M)	= 3 mH.
   	2L + 2M

   
   

   Correct Option: A

   As, 2L + 2M = 12
   2L – 2M = 4
   Then, L = 4
   ⇒ M = 2
   To get maximum value in parallel connection,
   

   Lmax =	(L + M)(L + M)	= 3 mH.
   	2L + 2M

   
   

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63. A resistance ‘R’ Ω and inductance of ‘L’ H are connected across 240 V, Hz supply. Power dissipated in the circuit is 300 W and the voltage across R is 100 V. In order to improve the power fact or to unity, the capacitor that is to be connect ed in series should have a val ue of _______ μF

1. 1. 11.5 μF.

   
   
   

   2. 13.7 μF.

   
   
   

   3. 12.5 μF.

   
   
   

   4. 43.7 μF.

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   	V2	= 300
   	R

   
   

   ⇒ R =	1002	=	100	Ω
   	300		3

   
   

   ⇒ I =	100				= 3 amp.
   		100
   		3

   
   For p.f. to be equal to unity,
   

   I(ωL) = I			1		= VL
   			ωC

   
   

   ⇒	1	= √240² - 100² = 218.17 Volt
   	ωC

   
   

   ⇒ C =	3	= 43.7 × 10-6 = 43.7 μF
   	2π × 50 × 218.17

   
   

   Correct Option: D

   
   

   	V2	= 300
   	R

   
   

   ⇒ R =	1002	=	100	Ω
   	300		3

   
   

   ⇒ I =	100				= 3 amp.
   		100
   		3

   
   For p.f. to be equal to unity,
   

   I(ωL) = I			1		= VL
   			ωC

   
   

   ⇒	1	= √240² - 100² = 218.17 Volt
   	ωC

   
   

   ⇒ C =	3	= 43.7 × 10-6 = 43.7 μF
   	2π × 50 × 218.17

   
   

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64. In the network shown in the figure, the effective resistance faced by the voltage source is _____ Ω .
    
    

1. 1. 3 Ω

   
   
   

   2. 7 Ω

   
   
   

   3. 9 Ω

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Current through the 4-ohm resistor = i -	i	=	3i
   	4		4

   
   

   Therefore, voltage drop across it =			3i		× 4 = 3i which must equal V.
   			4

   
   Thus effective resistance faced by the voltage source is 3 Ω .
   

   Correct Option: A

   Current through the 4-ohm resistor = i -	i	=	3i
   	4		4

   
   

   Therefore, voltage drop across it =			3i		× 4 = 3i which must equal V.
   			4

   
   Thus effective resistance faced by the voltage source is 3 Ω .
   

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65. Two coils in differential connection have self inductance of 2 mH and 4 mH and a mutual inductance of 0.15 mH. The equivalent inductance of the combination is _____ mH
    

1. 1. 1.7 mH

   
   
   

   2. 3.0 mH

   
   
   

   3. 5.7 mH

   
   
   

   4. 1.7 mH

   
   
   

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1. View Hint View Answer Discuss in Forum

   When two inductors are connected in series, the effective inductance
   L_eff = L₁ + L₂ ± 2M
   In this case, L_eff = L₁ + L₂ – 2M
   = 2 + 4 – 2 × 0.15 = 5.7 mH.

   Correct Option: C

   When two inductors are connected in series, the effective inductance
   L_eff = L₁ + L₂ ± 2M
   In this case, L_eff = L₁ + L₂ – 2M
   = 2 + 4 – 2 × 0.15 = 5.7 mH.

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