216. The time constant of the network shown below is given by—
     
     
     

1. 1. (1)

   
   
   

   2. (2)

   
   
   

   3. (3)

   
   
   

   4. (4)

   
   
   

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1. View Hint View Answer Discuss in Forum

   so, τ =	L
   	R3 + ( R1 + R2 / R1 + R2)

   
   

   Correct Option: A

   so, τ =	L
   	R3 + ( R1 + R2 / R1 + R2)

   
   

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217. The steady state current through the 1H inductor in the circuit shown in the given figure is—
     
     
     

1. 1. Zero

   
   
   

   2. 3A

   
   
   

   3. 5A

   
   
   

   4. 6A

   
   
   

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1. View Hint View Answer Discuss in Forum

   At steady state condition inductor will short
   Hence alternative (B) is the correct choice.
   
   

   
   

   Correct Option: B

   At steady state condition inductor will short Hence alternative (B) is the correct choice.

   
   

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218. For the circuit shown below, the switch is closed at t = 0. The current through the capacitor decreases exponentially with a time constant—
     
     
     

1. 1. 0·5s

   
   
   

   2. 1s

   
   
   

   3. 2s

   
   
   

   4. 10s

   
   
   

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1. View Hint View Answer Discuss in Forum

   τ = RC = Req. C = (1 || 1) × 1f =	1	× 1 = ·5S
   	2

   Correct Option: A

   τ = RC = Req. C = (1 || 1) × 1f =	1	× 1 = ·5S
   	2

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219. For the circuit shown below, C₁ = C₂ = 2F and the capacitor C₁ has a voltage of 20V when S is open. If the switch S is closed at t = 0, the voltage V_c2 will be a—
     
     
     

1. 1. Fixed voltage of 20V

   
   
   

   2. Fixed voltage of 10V

   
   
   

   3. Fixed voltage of – 10V

   
   
   

   4. Sinusoidal voltage

   
   
   

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1. View Hint View Answer Discuss in Forum

   Applying KVL in s-domain
   

   	LS +	2		I(s) =	20
   		2S			s

   
   

   	LS +	1		I(s) =	20
   		S			s

   
   

   I(s) =	20
   	LS2 + 1

   
   

   VC2(s) =	1	I(s) =	1	.	20
   	SC2		SC2		LS2 + 1

   
   Here A + B are constant after taking inverse laplace V_C2 (t) will be sinusoidal voltage.
   Hence alternative (D) is the correct choice.

   
   

   Correct Option: D

   Applying KVL in s-domain
   

   	LS +	2		I(s) =	20
   		2S			s

   
   

   	LS +	1		I(s) =	20
   		S			s

   
   

   I(s) =	20
   	LS2 + 1

   
   

   VC2(s) =	1	I(s) =	1	.	20
   	SC2		SC2		LS2 + 1

   
   Here A + B are constant after taking inverse laplace V_C2 (t) will be sinusoidal voltage.
   Hence alternative (D) is the correct choice.

   
   

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220. A resistor R of 1Ω and two inductors L₁ and L₂ of inductances 1H and 2H respectively, are connected in parallel. At some time, the currents through L₁ and L₂ are 1A and 2A. The current through R at time t = ∞ will be—

1. 1. Zero

   
   
   

   2. 1A

   
   
   

   3. 2A

   
   
   

   4. 3A

   
   
   

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1. View Hint View Answer Discuss in Forum

   at t = ∞ inductor will short circuited and because of parallel connection resistor will also be short circuited and current through R will be zero.
   

   Correct Option: A

   at t = ∞ inductor will short circuited and because of parallel connection resistor will also be short circuited and current through R will be zero.
   

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