6. In the circuit shown in the given figure, switch K is closed at t = 0. The circuit was initially relaxed. Which one of the following sources of v(t) will produce maximum current at t = 0⁺ ?
   
   

1. 1. Unit step

   
   
   

   2. Unit impulse

   
   
   

   3. Unit ramp

   
   
   

   4. Unit step plus unit ramp

   
   
   

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1. View Hint View Answer Discuss in Forum

   For unit step source,
   

   I(t) =	V		1 - e(-R / L)t
   	R

   
   At t = 0, I (t) = 0
   

   For ramp, inputI (t) I(t) =	V		tUt -		1 - e(-R / L)t
   	R

   
   At t = 0, I(t) = 0
   

   For impulse input, I(t) =	V	e(-R / L)t
   	R

   
   

   Then at t = 0, I(t) =	V	Maximum
   	R

   
   

   Correct Option: B

   For unit step source,
   

   I(t) =	V		1 - e(-R / L)t
   	R

   
   At t = 0, I (t) = 0
   

   For ramp, inputI (t) I(t) =	V		tUt -		1 - e(-R / L)t
   	R

   
   At t = 0, I(t) = 0
   

   For impulse input, I(t) =	V	e(-R / L)t
   	R

   
   

   Then at t = 0, I(t) =	V	Maximum
   	R

   
   

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7. A battery charger can drive a current of 5 A into a 1 ohm resistance connected at its output teminals. If it is able to charge an ideal 2 V battery at 7 A rate, then its Thevenin’s equivalent will be
   
   

1. 1. 7.5 V in series with 0.5 ohm

   
   
   

   2. 12 V in series with 1.5 ohms

   
   
   

   3. 7.5 V in parallel with 0.5 ohm

   
   
   

   4. 1.25 V in parallel with 1.5 ohms

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   From Fig. (a)
   V = I (1 + r)
   ⇒ V = 5 (1 + r) ...(i)
   From Fig. (b),
   V – 2 = 7 r ...(ii)
   Adding equations (i), and (ii), we get
   V = 12 volts and, r = 1.5 Ω
   

   Correct Option: B

   
   From Fig. (a)
   V = I (1 + r)
   ⇒ V = 5 (1 + r) ...(i)
   From Fig. (b),
   V – 2 = 7 r ...(ii)
   Adding equations (i), and (ii), we get
   V = 12 volts and, r = 1.5 Ω
   

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8. In the circuit shown in the given figure, steadystate was reached when the switch S was open. The switch was closed at t = 0. The initial value of the current through the capacitor 2C is 12 Amp
   
   

1. 1. zero

   
   
   

   2. 1 amp

   
   
   

   3. 2 amp

   
   
   

   4. 3 amp

   
   
   

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1. View Hint View Answer Discuss in Forum

   Initial voltage on 2C = 12 volts
   

   Then,discharging current, Id(t) =	d						2C.12 e(-t.3 / 8C)
   		dt

   
   

   Then , Id(t) t = 0| =		-24C ×	3	. e(-t.3 / 8C)
   			8C

   
   = -9 amp
   

   Now, charging current, Ic(t) =	d		24C		1 - e(-3t / 8C)
   	dt

   
   = -9 amp
   At t = 0, I_c(t) = 9 amp.
   Hence, at t = 0
   total current I(t) = I_c(t) + I_d(t) = 0 amp.
   

   Correct Option: A

   Initial voltage on 2C = 12 volts
   

   Then,discharging current, Id(t) =	d						2C.12 e(-t.3 / 8C)
   		dt

   
   

   Then , Id(t) t = 0| =		-24C ×	3	. e(-t.3 / 8C)
   			8C

   
   = -9 amp
   

   Now, charging current, Ic(t) =	d		24C		1 - e(-3t / 8C)
   	dt

   
   = -9 amp
   At t = 0, I_c(t) = 9 amp.
   Hence, at t = 0
   total current I(t) = I_c(t) + I_d(t) = 0 amp.
   

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9. A resistor R is connected to a voltage source V_s having an internal resistance R_s . A voltmeter of resistance Rm is connected across the terminals of the resistor R. The voltmeter will read a voltage of

1. 1. 	Vs Rs Rm
      	Rs Rm + Rs R + R Rm

   
   
   

   2. 	Vs R
      	R + Rs

   
   
   

   3. 	Vs Rm
      	R Rs + Rm

   
   
   

   4. 	Vs R Rm
      	Rs R + Rs Rm + R Rm

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   The figure will be as drawn below :
   
   Voltage measured across R = I_R. R
   

   = R × I ×	Rm
   	R + Rm

   
   

   =	RRm	×	Vs
   	R + Rm			Rs +	Rm R
   					Rm + R

   
   

   =	Vs.RRm
   	RsR + RsRm + RRm

   Correct Option: D

   The figure will be as drawn below :
   
   Voltage measured across R = I_R. R
   

   = R × I ×	Rm
   	R + Rm

   
   

   =	RRm	×	Vs
   	R + Rm			Rs +	Rm R
   					Rm + R

   
   

   =	Vs.RRm
   	RsR + RsRm + RRm

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10. Which one of the following impedance values of load will cause maximum power to be transferred to the load for the network shown in the given figure?
    
    

1. 1. (2 + j2) Ω

   
   
   

   2. (2 – j2) Ω

   
   
   

   3. – j2 Ω

   
   
   

   4. 2 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   Z_TH (impedance looking left through terminal AB)
   

   = 2j +	(2 + 2j).(-2j)
   	2 + 2j - 2j

   
   = 2j + 2 – 2j = 2Ω
   For maximum power, Z_TH = Z_L
   

   Correct Option: D

   Z_TH (impedance looking left through terminal AB)
   

   = 2j +	(2 + 2j).(-2j)
   	2 + 2j - 2j

   
   = 2j + 2 – 2j = 2Ω
   For maximum power, Z_TH = Z_L
   

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