281. In a series RLC circuit, the locus of the tip of the admittance phasor in the complex plane as the frequency is varied, is—

1. 1. A semicircle in the upper half of the G-B plane having the centre at
      

      		1	‚ 0		and radius	1
      		R				R

   
   
   

   2. A circle in the upper half of the G-B plane having the centre at
      

      			1	‚ 0		and radius	1
      			(2R)				(2R)

   
   
   

   3. A semicircle in the bottom half of the G-B plane having the centre at
      

      			–1	‚ 0		and radius	1
      			(2R)				(2R)

   
   
   

   4. A semicircle in the upper half of the G-B plane having the centre at
      

      	–	1	‚ 0		and radius	1
      		R				R

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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282. Which one of the following statements is not correct for the circuit shown at resonant frequency?
     

1. 1. The current is maximum

   
   
   

   2. The equivalent impedance is real

   
   
   

   3. The inductive and capacitive reactances are equal in magnitude

   
   
   

   4. The quality factor equals to	1		C
      	R		L

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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283. A parallel circuit has two branches. In one branch, R and L are in series and in the other branch, R and C are in series. The circuit will exhibit unity power factor when

1. 1. R =	L
      	C

   
   
   

   2. R = LC

   
   
   

   3. R =	C
      	L

   
   
   

   4. R =	L
      	C

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that qualify factor Q is given by relation
   Q = ωCR
   

   where    ω =	1
   	√LC

   
   At    Q = 1
   

   1 = ωCR =	1	CR
   	√LC

   
   

   or    R =	√	L
   		C

   
   Hence alternative (A) is correct choice.

   Correct Option: A

   We know that qualify factor Q is given by relation
   Q = ωCR
   

   where    ω =	1
   	√LC

   
   At    Q = 1
   

   1 = ωCR =	1	CR
   	√LC

   
   

   or    R =	√	L
   		C

   
   Hence alternative (A) is correct choice.

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284. A step voltage is applied to the circuit shown below. What is the transient current response of the circuit?
     

1. 1. Undamped sinusoidal

   
   
   

   2. Over-damped

   
   
   

   3. Under-damped

   
   
   

   4. Critically damped

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit
   
   Given
   

   Vi = u(t) then Vi(s) =	1
   	S

   
   Applying KVL to the input loop, we get
   

   Vit = i(t) R + L	di(t)	+	1	∫i(t) dt
   	dt		C

   
   Taking laplace transform both side, we get
   

   1	=		R + LS +	1		I(S)
   S				CS

   
   or
   

   1	=	RCS + LCS2 + 1
   S		S

   
   The characteristic equation RCS + LCS² + 1 = 0
   
   

   or S2 +	R	S +	1	= 0
   	L		LC

   
   

   2ξωn =	R
   	L

   
   and

   ω2n =	L
   	LC

   
   Now,

   ξ =	R
   	2ωn

   
   

   =	R
   	L2	√	1
   			LC

   
   Putting the given values of R, L and C, we get
   ξ = 1
   Since, ξ = 1
   Thus, the transient current response of the circuit is critically damped. Hence alternative (D) is the correct choice.

   Correct Option: C

   The given circuit
   
   Given
   

   Vi = u(t) then Vi(s) =	1
   	S

   
   Applying KVL to the input loop, we get
   

   Vit = i(t) R + L	di(t)	+	1	∫i(t) dt
   	dt		C

   
   Taking laplace transform both side, we get
   

   1	=		R + LS +	1		I(S)
   S				CS

   
   or
   

   1	=	RCS + LCS2 + 1
   S		S

   
   The characteristic equation RCS + LCS² + 1 = 0
   
   

   or S2 +	R	S +	1	= 0
   	L		LC

   
   

   2ξωn =	R
   	L

   
   and

   ω2n =	L
   	LC

   
   Now,

   ξ =	R
   	2ωn

   
   

   =	R
   	L2	√	1
   			LC

   
   Putting the given values of R, L and C, we get
   ξ = 1
   Since, ξ = 1
   Thus, the transient current response of the circuit is critically damped. Hence alternative (D) is the correct choice.

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285. For the circuit shown below, calculate I_s (Given that current in the galvanometer is zero)—
     

1. 1. 1 mA

   
   
   

   2. 2 mA

   
   
   

   3. 3 mA

   
   
   

   4. 4 mA

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Given that current in the galvanometer is zero. It mean potential at point B and C is equal.
   So, voltage at point
   B = 5V
   Current in branch
   

   BD =	5V	= 1mA
   	5K

   
   Therefore voltage at terminal
   A = 10K × 1mA + 5K × 1mA
   = 15 V
   Current in branch
   

   AC =	VA −VC	=	15 − 5	= 2mA.
   	5K		5K

   
   Thus, I_s = 1 + 2 = 3 mA
   Hence alternative (C) is the correct choice.

   Correct Option: C

   
   Given that current in the galvanometer is zero. It mean potential at point B and C is equal.
   So, voltage at point
   B = 5V
   Current in branch
   

   BD =	5V	= 1mA
   	5K

   
   Therefore voltage at terminal
   A = 10K × 1mA + 5K × 1mA
   = 15 V
   Current in branch
   

   AC =	VA −VC	=	15 − 5	= 2mA.
   	5K		5K

   
   Thus, I_s = 1 + 2 = 3 mA
   Hence alternative (C) is the correct choice.

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