Network Elements and the Concept of Circuit

Network Elements and the Concept of Circuit

Easy Questions

Network Theory

Network Elements and the Concept of Circuit
Network theory miscellaneous
  1. The Thevenin equivalent across AA′ is given by—










  1. View Hint View Answer Discuss in Forum

    Calculation for Rth

    Rth = 15 || 5 =
    15 × 5
    15 + 5


    or Rth =
    75
    		 =
    15
    		 Ω
    204

    Calculation for Vth
    I = 20 ×
    5
    		= 5 amp.
    15 + 5


    Vth = I × 5 = 5 × 5 = 25 V

    Correct Option: C

    Calculation for Rth

    Rth = 15 || 5 =
    15 × 5
    15 + 5


    or Rth =
    75
    		 =
    15
    		 Ω
    204

    Calculation for Vth
    I = 20 ×
    5
    		= 5 amp.
    15 + 5


    Vth = I × 5 = 5 × 5 = 25 V

  1. The voltage marked Vx is given by—










  1. View Hint View Answer Discuss in Forum

    Apply superposition theorem to solve this problem
    Case I. Take 8A current source the circuit becomes

    current in 2 Ω resistance, i.e.

    I 1 = 8 ×
    5
    		 =
    40
    		 amp. → (i) (A → B)
    5 + 27

    Case II. Take 20 V voltage source: the circuit becomes
    Note: Here the resistance connected in the arm CD is the extra redundant element because voltage their resistance in independent.

    I = I2 =
    20
    		 =
    20
    		 amp
    5 + 27

    which is also flowing in 2 Ω resistance. (A → B) so, the net current in the 2 resistance
    =
    40
    		 +
    20
    		 =
    60
    		 (∵ I = I1 + I2)
    777

    Voltage (Vx) = 2 ×
    60
    		 =
    120
    		 V
    77

    Correct Option: B

    Apply superposition theorem to solve this problem
    Case I. Take 8A current source the circuit becomes

    current in 2 Ω resistance, i.e.

    I 1 = 8 ×
    5
    		 =
    40
    		 amp. → (i) (A → B)
    5 + 27

    Case II. Take 20 V voltage source: the circuit becomes
    Note: Here the resistance connected in the arm CD is the extra redundant element because voltage their resistance in independent.

    I = I2 =
    20
    		 =
    20
    		 amp
    5 + 27

    which is also flowing in 2 Ω resistance. (A → B) so, the net current in the 2 resistance
    =
    40
    		 +
    20
    		 =
    60
    		 (∵ I = I1 + I2)
    777

    Voltage (Vx) = 2 ×
    60
    		 =
    120
    		 V
    77

  1. In the circuit V2 = 2 V and I1 = 2 amp. The value of Is is given by—








  1. View Hint View Answer Discuss in Forum

    This problem can be easily solve by using superposition theorem
    Case I. Take current source, the circuit becomes

    I1 = Is ×
    (1 + 1 || 1)
    		 =
    Is . 3 / 2
    (2 + 1 + 1 || 1)2 + 3 / 2

    =
    3 Is
    		amp.
    7

    Case II. Take voltage source, the circuit becomes

    I =
    V
    Req

    where Req = equivalent resistance as seen from the terminal AB
    R eq = 3 || 1 + 1 =
    3
    		 + 1 =
    7
    44

    V = 2V
    I =
    2
    		 =
    8
    		 amp.
    7/47

    I 2 = 1 ×
    1
    		 =
    8
    		.
    1
    		 =
    2
    1 + 3747

    Now, the net current I = I1 + I2
    I =
    3Is
    		 +
    2
    77

    But given that
    I = 2 =
    3Is
    		 +
    2
    77
    or 7 × 2 = 3 Is + 2
    or 14 – 2 = 3 Is
    or Is = 4 amp.

    Correct Option: B

    This problem can be easily solve by using superposition theorem
    Case I. Take current source, the circuit becomes

    I1 = Is ×
    (1 + 1 || 1)
    		 =
    Is . 3 / 2
    (2 + 1 + 1 || 1)2 + 3 / 2

    =
    3 Is
    		amp.
    7

    Case II. Take voltage source, the circuit becomes

    I =
    V
    Req

    where Req = equivalent resistance as seen from the terminal AB
    R eq = 3 || 1 + 1 =
    3
    		 + 1 =
    7
    44

    V = 2V
    I =
    2
    		 =
    8
    		 amp.
    7/47

    I 2 = 1 ×
    1
    		 =
    8
    		.
    1
    		 =
    2
    1 + 3747

    Now, the net current I = I1 + I2
    I =
    3Is
    		 +
    2
    77

    But given that
    I = 2 =
    3Is
    		 +
    2
    77
    or 7 × 2 = 3 Is + 2
    or 14 – 2 = 3 Is
    or Is = 4 amp.

  1. The current I as marked in the figure is—










  1. View Hint View Answer Discuss in Forum

    On applying superposition theorem.
    Case I. Taken current source in this situation the 6A circuit becomes.

    I =
    6 × 1
    		2 (A → B)
    2 + 1
    Case II. Taken voltage source: In this situation the circuit becomes I 1 O.C. 3V 1Ω 3Ω 4Ω A B 2Ω − + Note: Here 3 Ω resistance is a extra element because voltage at node B is independent of the resistance 3 Ω.
    I1 =
    3
    		= 1 amp. (B → A)
    2 + 1

    The net current in 2 Ω resistance
    i = I – I1 = 2 – 1 = 1 amp. (A → B)


    Correct Option: C

    On applying superposition theorem.
    Case I. Taken current source in this situation the 6A circuit becomes.

    I =
    6 × 1
    		2 (A → B)
    2 + 1
    Case II. Taken voltage source: In this situation the circuit becomes I 1 O.C. 3V 1Ω 3Ω 4Ω A B 2Ω − + Note: Here 3 Ω resistance is a extra element because voltage at node B is independent of the resistance 3 Ω.
    I1 =
    3
    		= 1 amp. (B → A)
    2 + 1

    The net current in 2 Ω resistance
    i = I – I1 = 2 – 1 = 1 amp. (A → B)


  1. A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled, then the voltage across the resistor is—








  1. View Hint View Answer Discuss in Forum

    Let the network contains two resistors in

    I =
    V
    		or V = 2 IR ....(1)
    R

    If the value of resistance increases to two times
    I becomes
    1
    		and R becomes 2R.
    2

    so new voltage
    V′ = 2.
    1
    		(2R) = 2 IR = V
    2

    Hence, alternative (D) is correct.


    Correct Option: D

    Let the network contains two resistors in

    I =
    V
    		or V = 2 IR ....(1)
    R

    If the value of resistance increases to two times
    I becomes
    1
    		and R becomes 2R.
    2

    so new voltage
    V′ = 2.
    1
    		(2R) = 2 IR = V
    2

    Hence, alternative (D) is correct.