16. The impedance seen by the source in the given circuit is
    
    

1. 1. (0.54 + j0.313) ohms

   
   
   

   2. (4 – j2) ohms

   
   
   

   3. (4.54 – j1.69) ohms

   
   
   

   4. (4 + j2) ohms

   
   
   

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1. View Hint View Answer Discuss in Forum

   Z'L = 10 ∠ 30		1		2
   		4

   
   = (0.54 + j0.31) Ω
   Total impedance = (4.54 – j 1.69) Ω.

   Correct Option: C

   Z'L = 10 ∠ 30		1		2
   		4

   
   = (0.54 + j0.31) Ω
   Total impedance = (4.54 – j 1.69) Ω.

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17. Consider star network shown in the given figure. The resistance between terminals A and B with C open is 6 Ω, between terminals B and C with A open is 11 Ω, and between terminals C and A with B open is 9 Ω. Then
    
    

1. 1. R_A = 4 Ω , R_B = 2 Ω, R_C = 5 Ω

   
   
   

   2. R_A = 2 Ω , R_B = 4 Ω, R_C = 7 Ω

   
   
   

   3. R_A = 3 Ω , R_B = 3 Ω , R_C = 4 Ω

   
   
   

   4. R_A = 5 Ω, R_B = 1 Ω , R_C = 10 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : R_A + R_B = 6 with C open
   R_B + R_C = 11 with A open
   R_C + R_A = 9
   

   ∴ RA + RB + RC =	26	= 13
   	2

   
   ⇒ R_A = 2 Ω
   
   

   Correct Option: B

   Given : R_A + R_B = 6 with C open
   R_B + R_C = 11 with A open
   R_C + R_A = 9
   

   ∴ RA + RB + RC =	26	= 13
   	2

   
   ⇒ R_A = 2 Ω
   
   

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18. Current i(t), through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4sin (100t + 45°) + 4sin (300 t + 60°) A RMS value of the current and power dissipated in the circuit respectively are
    

1. 1. √41 A , 410 W
      

   
   
   

   2. √35 A , 350 W
      

   
   
   

   3. 5 A , 250 W
      

   
   
   

   4. 11 A , 1210 W

   
   
   

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1. View Hint View Answer Discuss in Forum

   i = 3+ 4 sin (1000 + 45°) + 4 sin (300 + 60°) amp
   
   Power = i² R(RMS)
   = 25 × 10 = 250 W

   Correct Option: C

   i = 3+ 4 sin (1000 + 45°) + 4 sin (300 + 60°) amp
   
   Power = i² R(RMS)
   = 25 × 10 = 250 W

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19. A circuit consisting of a 1Ω resistor and a 2 F capacitor in series is excited from a voltage source with the voltage expressed as 3e ^{– 4}, as shown in the given figure. If i(0^– ) and V_c (0^– ) both are zero, then values of i(0⁺) and i(∞) will be respectively
    
    

1. 1. 3 A and 1.5 A

   
   
   

   2. 1.5 A and zero

   
   
   

   3. 3 A and zero

   
   
   

   4. 1.5 A and 3A

   
   
   

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1. View Hint View Answer Discuss in Forum

   I(s)		R +	1		=	3
   			Cs			s + 1

   
   

   ⇒ I(s) =	3Cs	=	6s
   	(s + 1)(RCs + 1)		(s + 1)(2s + 1)

   
   

   Then , i(t) = Z-1 I(s) = Z-1	3s
   	(s + 1){s + (1 / 2)}

   
   
   i (t) = 6e^{– t} – 3e^{– t / 2}
   ∴ i (∞) = 3A, I(0) = 0
   

   Correct Option: C

   I(s)		R +	1		=	3
   			Cs			s + 1

   
   

   ⇒ I(s) =	3Cs	=	6s
   	(s + 1)(RCs + 1)		(s + 1)(2s + 1)

   
   

   Then , i(t) = Z-1 I(s) = Z-1	3s
   	(s + 1){s + (1 / 2)}

   
   
   i (t) = 6e^{– t} – 3e^{– t / 2}
   ∴ i (∞) = 3A, I(0) = 0
   

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20. Viewed from the terminal AB, following circuit shown in the figure can be reduced to an equivalent circuit of a single voltage source in series with a single resistor with which of the following parameters?
    
    

1. 1. 4 volts source in series with 10 Ω resistor

   
   
   

   2. 1 volt source in series with 2.4 Ω resistor

   
   
   

   3. 15 volts source in series with 2.4 Ω resistor

   
   
   

   4. 1 volt source in series with 10 Ω resistor

   
   
   

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1. View Hint View Answer Discuss in Forum

   Applying Thevenin’s theorem
   R_eq = 6 Ω | | 4 Ω = 2.4 Ω
   

   Vab = 10 - 6		15		= 1 V
   		10

   
   

   Correct Option: B

   Applying Thevenin’s theorem
   R_eq = 6 Ω | | 4 Ω = 2.4 Ω
   

   Vab = 10 - 6		15		= 1 V
   		10

   
   

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