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Electrical and Electronic Measurements Test 5
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Electrical and Electronic Measurements Test 5
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  • Question 1/10
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    The output voltage for the shown voltage divider network is

    Solutions

    quivalent circuit across the point V1 = R || 2R = 2R/3

    By using voltage division,

    V1=Vin(2R3R+2R3)=25V

    V0=V1(RR+R)=15V

  • Question 2/10
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    What is the maximum voltage (in V) that can be measured with the above potentiometer if the standardization is done with a 1.02 V standard cell, keeping the jockey at 510 cm during standardization?

    Solutions

    The length of the slide wire = 10 m = 1000 cm

    The potentiometer is standardized at 510 cm.

    The voltage corresponding to 510 cm = 1.02 V

    Voltage corresponding to 1 cm =1.02510=2×103V

    Maximum voltage that can be measured = length of slide wire × voltage per cm

    = 1000 × 2 × 10-3 = 2 V

  • Question 3/10
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    A voltmeter of 1 kΩ resistance and a mili ammeter of 1 Ω resistance are used to measure a unknown resistance by voltmeter-ammeter method. If the voltmeter is connected directly in series with the unknown resistance. If the voltmeter reads 50 V and milli ammeter reads 100 mA, then the magnitude of percentage error in the value of measured resistance is 

    Solutions

    Voltmeter reading (V) = 50 V

    Milli ammeter reading (I) = 100 mA

    Measruedresistance(Rm)=VI=50100×103=500Ω 

    Ammeter is connected in to load side. So, the error will be voltage drop across ammeter.

    True voltage drop across resistance = Voltmeter reading – Voltage drop across ammeter

    = 50 – (1) (100 × 10-3) = 49.9 V

    Truevalueofresistance(RT)=49.9100×103=499Ω 

    Percentageerror=RmRTRT×100 

    =500499499×100=0.2% 

  • Question 4/10
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    Which of the following bridges can be balanced?

    (Assume all components to be ideal and their values cannot be made zero or infinite)
    Solutions

    Bridge A: At bridge balance condition,

    (1jωC)R2=R1(jωL)

    R2=ω2R1LCω2=R2R1LC

    The above equation is not possible.

    Therefore, bridge A cannot be balanced.

    Bridge B:

    At bridge balance condition,

    (1jωC1)(1jωC2)=R1R2

    2R1R2 C1C2 = 1

    ω2=1R1R2C1C2

    The above equation is not possible.

    Therefore, bridge B cannot be balanced.

    Bridge C:

    At bridge balance condition,

    R4(R1(1jωC1)R1+1jωC1)=R3(jωL+R2)

    R4(R11+jωC1R1)=R3(R2+jωL)

    R1R4 = R3 (R2 + jωL) (1 + jωR1C1)

    R1R4 = R2R3 + jω R1C1R2R3 + jωLR3 – ω2L R1C1R3

    By comparing both sides,

    R1 C1 R2 R3 + LR3 = 0 L = -R1C1R2

    The above equation is not possible.

    Therefore, bridge C cannot be balanced.

    Bridge D:

    At bridge balance condition,

    R4(R1(1jωC1)R1+1jωC1)=R3(R2+1jωC2)

    R1R4(11+jωR1C1)=R3(1+jωC2R2)jωC2  

    jωR1R4C2 = R3 (1 + jωC2R2) (1 + jωR1C1)

    jωR1R4C2 = R3 + jωR3C2R2 + jωR1C1R3 – ω2R1R2R3C1C2

    By comparing both the sides,

    R3=ω2R1R2R3C1C2ω2=1R1R2C1C2

    R1R4C2 = R3C2R2 + R1C1R3     

    The above conditions can be true by selecting appropriate values of R1, R2, R3, R4, C1 and C2.

    Therefore, the bridge D can be balanced.

  • Question 5/10
    1 / -0

    A high voltage capacitor is investigated using the Schering bridge shown above. At balance, R3 = 1 kΩ, R4 = 10 kΩ, C1 = 0.1 μF, C3 = 1200 pF.

    The dissipation factor of the unknown capacitor if supply frequency is 400 Hz is

    Solutions

    At bridge balance condition,

    (1jωC1)(R4)=(1jωCx+Rx)(R3(1jωC3)R3+1jωC3)

    (R4jωC1)=(1+jωCxRxjωCx)(R31+jωR3C3)

    R4Cx (1 + jωR3C3) = R3C1 (1 + jω Cx Rx)

    By comparing both sides

    R4Cx=R3C1Cx=R3C1R4

    R3C3=CxrxRx=R3C3Cx=R3C3×R4R3C1

    Rx=C3C1R4

    Dissipation factor = ωCxRx = ωR3C3

    = 2πf R3C3

    = 2π × 400 × 1 × 103 × 1200 × 10-12

    = 0.003   

  • Question 6/10
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    The Hay bridge in the above figure operating at a supply frequency of 5002πHz, is balanced when the components are R1 = 1 kΩ, C3 = 0.2 μF, R3 = 10 Ω and R4 = 500 Ω. Unknown inductor coil is in the second arm (Zx) of the bridge.

    The value of inductance Lp in Henry is _______

    Solutions

    At bridge balance condition, Z1Z4 = Z2Z3

    R1R4=(Rp(jωLp)Rp+jωLp)(R3+1jωC3)

    R1R4(Rp+jωLp)=RpLpC3(1+jωR3C3)

    R1R4Rp + jω R1R4 LpRpLpC3+jωR3RpLp

    By comparing on both sides,

    R1R4Rp=RpLpC3

    Lp = R1R4 C3

    Lp = 1 × 103 × 500 × 0.2 × 10-6 = 0.1 H

  • Question 7/10
    1 / -0

    Figure shows a balanced ac bridge excited by a voltage source of fixed frequency. The impedance of the unknown arm Z is

    Solutions

    Let Z = (R4 + jωL4)

    At bridge balance condition

    R1R2 = (R3 + jωL3) (R4 + jωL4)

    R1R2 = R3R4 – ω2L3L4 + jω(L3R4 + R3L4)

    The above bridge can’t be balanced as the term (L3R4 + R3L4) can’t be zero.

    So, the unknown impedance Z should be capacitive in nature.

    Let z = R4 and C4 in series

    Z=R4=1jωC4

    At bridge balance condition,

    R1R2=(R3+jωL3)(R4+1jωC4)

    R1R2=R3R4+L3C4+j(ωL3R4R3ωC4)

    R1R2=R3R4+L3C4 and ωL3R4=R3ωC4

    As source frequency is fixed, we need to vary L3 and R3 to obtain bridge balance.

    As R3 and L3 are present in the both equations, it is difficult to obtain balance.

    Let z = R4 and C4 in parallel.

    z=R4(1jωC4)R4+1jωC4=R41+jωR4C4

    At bridge balance condition,

    R1R2=(R3+jωL3)(R41+jωR4C4)

    ⇒ R1R2 + jωR1R2R4C4 = R3R4 + jωL3R4

    R4=R1R2R3andC4=L3R1R2

    Now it is easy to obtain bridge balance condition. So, the z is a parallel combination of R and C.
  • Question 8/10
    1 / -0

    Consider the unbalanced Wheatstone bridge shown in figure.

    The internal resistance of the galvanometer is 1 ohm. The galvanometer current is ___________ (in mA)

    Solutions

    Vth = Va - Vb

    =10(100100+100)10(90110+90)=0.5V 

    A screenshot of a cell phoneDescription automatically generated

    Rth = Rab = (110 || 90) + (100 || 100)

    = 99.5Ω

    A screenshot of a cell phoneDescription automatically generated

    Ig=0.5100.5=4.975mA

  • Question 9/10
    1 / -0

    A length of label is tested for insulation resistance by loss of charge method. An electrostatic voltmeter of infinite resistance is connected between the cable conductor such that joint capacitance is 500 μF. If the voltage falls from 400 V to 100 V in one minute then the insulation resistance of the cable will be _______ (in k Ω)
    Solutions

    Insulation resistance of the cable is,

    R=0.4343tClog(Vv)

    R=0.4343×60500×106log(400100)

    ⇒ R = 86.56 kΩ
  • Question 10/10
    1 / -0

    An unknown voltage source Eu (with negligible internal resistance) is connected to a potentiometer circuit as shown in the following figure. If the galvanometer current is 10 μA with the direction as indicated, then value of Eu is

    Solutions

    By applying KVL,

    -1.6 + 100 Iw + 500 Iw + 1000 (Iw + Ig) = 0

    ⇒ 1600 Iw + 1000 Ig = 1.6

    ⇒ 1600 Iw + 1000 (10 × 10-6) = 1.6

    ⇒ 1600 Iw + 0.01 = 1.6

    Iw=1.60.011600=9.9375×104A 

    By applying KVL,

    -Eu + Ig(100) + (Iw + Ig) (1000) = 0

    ⇒ Eu = 1000 Iw + 1100 Ig

    = 0.99375 + 0.011 = 1.00475 V

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