91. The percentage of power loss in the question 115—

1. 1. 5%

   
   
   

   2. 2.5%

   
   
   

   3. 10%

   
   
   

   4. 20%

   
   
   

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1. View Hint View Answer Discuss in Forum

   % loss
   

   =		Power dissipated	× 100 =	2.5 × 10–3	× 100
   		Max. energy		0.1

   
   = 2.5%

   Correct Option: B

   % loss
   

   =		Power dissipated	× 100 =	2.5 × 10–3	× 100
   		Max. energy		0.1

   
   = 2.5%

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92. What will be the power dissipated in R in above question?

1. 1. 2.5 × 10^–3 joule

   
   
   

   2. 25 × 10^–3 joule

   
   
   

   3. .25 × 10^–3 joule

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since energy will be maximum when
   sin 2πt = 1
   

   sin 2πt = sin	π
   	2

   
   

   t =	1	sec
   	4

   
   Power dissipated
   = I² R = (10^{– 4} sin 2πt)². 10⁶ = 10^{– 2} sin² 2πt
   Power dissipated
   

   =		1 / 2	(10–2 sin2 2πt) dt
   		0

   
   

   =		1 / 2	1	(1 – cos 4πt) dt
   		0	2

   
   = 2.5 × 10– 3 joule
   Note:

   Since the current flows in resistor for		1	sec during charging and	1	sec during discharging so
   		4		4

   
   

   t =		1	+	1	=	1	sec.
   		4		4		2

   
   

   ∴ Power dissipated from 0 to	1	sec .
   	2

   Correct Option: A

   Since energy will be maximum when
   sin 2πt = 1
   

   sin 2πt = sin	π
   	2

   
   

   t =	1	sec
   	4

   
   Power dissipated
   = I² R = (10^{– 4} sin 2πt)². 10⁶ = 10^{– 2} sin² 2πt
   Power dissipated
   

   =		1 / 2	(10–2 sin2 2πt) dt
   		0

   
   

   =		1 / 2	1	(1 – cos 4πt) dt
   		0	2

   
   = 2.5 × 10– 3 joule
   Note:

   Since the current flows in resistor for		1	sec during charging and	1	sec during discharging so
   		4		4

   
   

   t =		1	+	1	=	1	sec.
   		4		4		2

   
   

   ∴ Power dissipated from 0 to	1	sec .
   	2

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93. Determine the maximum energy stored in the capacitor C for the figure given below—
    
    
    

1. 1. 1 joule

   
   
   

   2. 10 joule

   
   
   

   3. 0.1 joule

   
   
   

   4. .001 joule

   
   
   

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1. View Hint View Answer Discuss in Forum

   Maximum energy stored in the capacitor is given by the relation
   

   =	1	CV2
   	2

   
   = 1 2 20 × 10^{– 6} (100 sin 2πt)²
   = 1 2 × 20 × 10^{– 6} × 10⁴ sin² 2πt
   = 0.1 sin² 2πt joule
   E_max = 0.1 joule
   (E_max when sin² 2πt = 1)

   Correct Option: A

   Maximum energy stored in the capacitor is given by the relation
   

   =	1	CV2
   	2

   
   = 1 2 20 × 10^{– 6} (100 sin 2πt)²
   = 1 2 × 20 × 10^{– 6} × 10⁴ sin² 2πt
   = 0.1 sin² 2πt joule
   E_max = 0.1 joule
   (E_max when sin² 2πt = 1)

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94. What will be output voltage across the capacitor for the given input shown below?
    
    
    

1. 1. 8 V

   
   
   

   2. 12 V

   
   
   

   3. 10 V

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Voltage across the capacitor =	1			i dt
   	C

   
   

   =	1		2ms	20 × 10–3 dt
   	5 × 10–6		0

   
   

   	1	20 × 10–3 × 2 × 10–3 dt
   	5 × 10–6

   
   

   =	40	= 8V
   	5

   Correct Option: A

   Voltage across the capacitor =	1			i dt
   	C

   
   

   =	1		2ms	20 × 10–3 dt
   	5 × 10–6		0

   
   

   	1	20 × 10–3 × 2 × 10–3 dt
   	5 × 10–6

   
   

   =	40	= 8V
   	5

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95. Find the Thevenin voltage and resistance for the given circuit shown below—
    
    
    

1. 1. R_th = 58.82 Ω, V_th = 117.64 V

   
   
   

   2. R_th = 117.64 Ω, V_th = 58.82 V

   
   
   

   3. R_th = 58.82 Ω, V_th = 58.82 V

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Calculation for R_th
   As there are many dependent sources are taking into account so in order to calculate the R_th let an imaginary current source say 2A is connected across the open terminal then R_th will be the ratio of
   

   		Vx	=	Vx
   		I		2

   
   
   On applying KCL at node N.
   

   		Vx - 1000 Ix	+	Vx	+	Vx	= 2 ....(i)
   		100		100		3000

   
   

   Also IX =	Vx	= I1....(ii)
   	3000

   
   Solving equations (i) and (ii), we get
   V_X = (58.82 × 2) V
   

   so, Rth =		Vx	=	58.82 × 2	= 58.82 Ω
   		2		2

   
   

   (∵ Rth =	Vx	= I1, where I = 2 amp {imaginary source})
   	I

   
   V_X = V_th = 58.82 × 2 = 117.6 V

   Correct Option: A

   Calculation for R_th
   As there are many dependent sources are taking into account so in order to calculate the R_th let an imaginary current source say 2A is connected across the open terminal then R_th will be the ratio of
   

   		Vx	=	Vx
   		I		2

   
   
   On applying KCL at node N.
   

   		Vx - 1000 Ix	+	Vx	+	Vx	= 2 ....(i)
   		100		100		3000

   
   

   Also IX =	Vx	= I1....(ii)
   	3000

   
   Solving equations (i) and (ii), we get
   V_X = (58.82 × 2) V
   

   so, Rth =		Vx	=	58.82 × 2	= 58.82 Ω
   		2		2

   
   

   (∵ Rth =	Vx	= I1, where I = 2 amp {imaginary source})
   	I

   
   V_X = V_th = 58.82 × 2 = 117.6 V

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