Electric circuits miscellaneous

Electric circuits miscellaneous

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Electric Circuits

Electric circuits miscellaneous
  1. The sinusoidal steady-state voltage gain of the network shown in the given figure will have magnitude equal to 0.707 at an angular frequency of









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    M(jω) =
    V2(jω)
    V1(jω)

    Correct Option: C



    M(jω) =
    V2(jω)
    V1(jω)

  1. For the circuit shown in the figure, the current I is









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    Redrawing the circuit, we have

    From the circuit, V = 2RI ....(i)
    and, V = (3R)4 + (4R). 1 = 16 R ....(ii)
    From equations (i) and (ii), we get I = 8A

    Correct Option: D

    Redrawing the circuit, we have

    From the circuit, V = 2RI ....(i)
    and, V = (3R)4 + (4R). 1 = 16 R ....(ii)
    From equations (i) and (ii), we get I = 8A

  1. If a resistance R of 1 Ω is connected across the terminals AB as shown in the given figure, then the current flowing through R will be










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    Shorting the voltage source and opening current source, we have ZTH = 1 Ω looking left into terminal AB.

    Now, for VTH, from the circuit, we have
    VAB = VTH = 1 volt
    Then, current through R

    =
    VAB
    		 =
    1
    		 = 0.54
    R + ZTH2

    Correct Option: B

    Shorting the voltage source and opening current source, we have ZTH = 1 Ω looking left into terminal AB.

    Now, for VTH, from the circuit, we have
    VAB = VTH = 1 volt
    Then, current through R

    =
    VAB
    		 =
    1
    		 = 0.54
    R + ZTH2

  1. In the given circuit, V1 = 40 V when R is 10 Ω . When R is zero, the value of V2 will be









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    Applying KCL at node (1), we have

    V1 - 60
    		 +
    V1
    		 +
    V1 - V2
    		 = 0
    1020R

    ⇒
    V1 - 60
    		 +
    V1
    		 +
    V1
    		 =
    V2
    1020RR

    ⇒ V2 =
    V1 - 60
    		 +
    V1
    		R + V2
    1020

    R = 0 gives, V2 = V1
    Also ,
    V1 - 60
    		 +
    V1
    		 = 0
    1020

    ⇒ V1 = 40 volts
    ∴ V2 = 0 V

    Correct Option: A

    Applying KCL at node (1), we have

    V1 - 60
    		 +
    V1
    		 +
    V1 - V2
    		 = 0
    1020R

    ⇒
    V1 - 60
    		 +
    V1
    		 +
    V1
    		 =
    V2
    1020RR

    ⇒ V2 =
    V1 - 60
    		 +
    V1
    		R + V2
    1020

    R = 0 gives, V2 = V1
    Also ,
    V1 - 60
    		 +
    V1
    		 = 0
    1020

    ⇒ V1 = 40 volts
    ∴ V2 = 0 V

  1. In the circuit shown in the given figure, if I1 = 1.5 A, then I2 will be









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    From the circuit, we have
    6 = 2I1 + 2I2
    ⇒ I2 = 1.5A

    Correct Option: C

    From the circuit, we have
    6 = 2I1 + 2I2
    ⇒ I2 = 1.5A