Electrical and Electronic Measurements Test 5

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Question 1/10
1 / -0

The output voltage for the shown voltage divider network is

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A
$V_{0} = \frac{1}{3} V$

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B
$V_{0} = \frac{1}{4} V$

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C
$V_{0} = \frac{1}{5} V$

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D
$\frac{1}{6} V$

Solutions

quivalent circuit across the point V₁ = R || 2R = 2R/3
By using voltage division,
$V_{1} = V_{i n} (\frac{\frac{2 R}{3}}{R + \frac{2 R}{3}}) = \frac{2}{5} V$
$V_{0} = V_{1} (\frac{R}{R + R}) = \frac{1}{5} V$
Question 2/10
1 / -0

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?

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A
2-2

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B
24-21

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C
27-26

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D
29-22

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 $= \frac{1.02}{510} = 2 \times 10^{- 3} V$
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

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A
0.19-0.21

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B
0.194-0.211

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C
0.197-0.216

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D
0.199-0.212

Solutions

Voltmeter reading (V) = 50 V
Milli ammeter reading (I) = 100 mA
$M e a s r u e d r e s i s t a n c e (R_{m}) = \frac{V}{I} = \frac{50}{100 \times 10^{- 3}} = 500 Ω$
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
$T r u e v a l u e o f r e s i s t a n c e (R_{T}) = \frac{49.9}{100 \times 10^{- 3}} = 499 Ω$
$P e r c e n t a g e e r r o r = \frac{R_{m} - R_{T}}{R_{T}} \times 100$
$= \frac{500 - 499}{499} \times 100 = 0.2 %$
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)

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A

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B

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C

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D

Solutions

Bridge A: At bridge balance condition,
$(\frac{1}{j ω C}) R_{2} = R_{1} (j ω L)$
$\Rightarrow R_{2} = - ω^{2} R_{1} L C \Rightarrow ω^{2} = - \frac{R_{2}}{R_{1} L C}$
The above equation is not possible.
Therefore, bridge A cannot be balanced.
Bridge B:
At bridge balance condition,
$(\frac{1}{j ω C_{1}}) (\frac{1}{j ω C_{2}}) = R_{1} R_{2}$
⇒ -ω²R₁R₂ C₁C₂ = 1
$\Rightarrow ω^{2} = - \frac{1}{R_{1} R_{2} C_{1} C_{2}}$
The above equation is not possible.
Therefore, bridge B cannot be balanced.
Bridge C:
At bridge balance condition,
$R_{4} (\frac{R_{1} (\frac{1}{j ω C_{1}})}{R_{1} + \frac{1}{j ω C_{1}}}) = R_{3} (j ω L + R_{2})$
$\Rightarrow R_{4} (\frac{R_{1}}{1 + j ω C_{1} R_{1}}) = R_{3} (R_{2} + j ω L)$
⇒ 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,
$R_{4} (\frac{R_{1} (\frac{1}{j ω C_{1}})}{R_{1} + \frac{1}{j ω C_{1}}}) = R_{3} (R_{2} + \frac{1}{j ω C_{2}})$
$\Rightarrow R_{1} R_{4} (\frac{1}{1 + j ω R_{1} C_{1}}) = \frac{R_{3} (1 + j ω C_{2} R_{2})}{j ω C_{2}}$
⇒ jωR₁R₄C₂ = R₃ (1 + jωC₂R₂) (1 + jωR₁C₁)
⇒ jωR₁R₄C₂ = R₃ + jωR₃C₂R₂ + jωR₁C₁R₃ – ω²R₁R₂R₃C₁C₂
By comparing both the sides,
$R_{3} = ω^{2} R_{1} R_{2} R_{3} C_{1} C_{2} \Rightarrow ω^{2} = \frac{1}{R_{1} R_{2} C_{1} C_{2}}$
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

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A
0.003

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B
0.002

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C
0.004

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D
0.001

Solutions

At bridge balance condition,
$(\frac{1}{j ω C_{1}}) (R_{4}) = (\frac{1}{j ω C_{x}} + R_{x}) (\frac{R_{3} (\frac{1}{j ω C_{3}})}{R_{3} + \frac{1}{j ω C_{3}}})$
$\Rightarrow (\frac{R_{4}}{j ω C_{1}}) = (\frac{1 + j ω C_{x} R_{x}}{j ω C_{x}}) (\frac{R_{3}}{1 + j ω R_{3} C_{3}})$
⇒ R₄C_x (1 + jωR₃C₃) = R₃C₁ (1 + jω C_x R_x)
By comparing both sides
$R_{4} C_{x} = R_{3} C_{1} \Rightarrow C_{x} = \frac{R_{3} C_{1}}{R_{4}}$
$R_{3} C_{3} = C_{x} r_{x} \Rightarrow R_{x} = \frac{R_{3} C_{3}}{C_{x}} = \frac{R_{3} C_{3} \times R_{4}}{R_{3} C_{1}}$
$\Rightarrow R_{x} = \frac{C_{3}}{C_{1}} R_{4}$
Dissipation factor = ωC_xR_x = 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 $\frac{500}{2 π} H z$ , 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 (Z_x) of the bridge.
The value of inductance L_p in Henry is _______

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A
0.09-0.11

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B
0.094-0.111

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C
0.097-0.116

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D
0.099-0.112

Solutions

At bridge balance condition, Z₁Z₄ = Z₂Z₃
$\Rightarrow R_{1} R_{4} = (\frac{R_{p} (j ω L_{p})}{R_{p} + j ω L_{p}}) (R_{3} + \frac{1}{j ω C_{3}})$
$\Rightarrow R_{1} R_{4} (R_{p} + j ω L_{p}) = \frac{R_{p} L_{p}}{C_{3}} (1 + j ω R_{3} C_{3})$
⇒ R₁R₄R_p + jω R₁R₄ L_p = $\frac{R_{p} L_{p}}{C_{3}} + j ω R_{3} R_{p} L_{p}$
⇒ By comparing on both sides,
$\Rightarrow R_{1} R_{4} R_{p} = \frac{R_{p} L_{p}}{C_{3}}$
⇒ L_p = R₁R₄ C₃
L_p = 1 × 10³ × 500 × 0.2 × 10^-6 = 0.1 H
Question 7/10
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Figure shows a balanced ac bridge excited by a voltage source of fixed frequency. The impedance of the unknown arm Z is

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A
an R and L in series

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B
an R and C in series

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C
an R and C in parallel

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D
an R and L in parallel

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
$Z = R_{4} = \frac{1}{j ω C_{4}}$
At bridge balance condition,
$R_{1} R_{2} = (R_{3} + j ω L_{3}) (R_{4} + \frac{1}{j ω C_{4}})$
$\Rightarrow R_{1} R_{2} = R_{3} R_{4} + \frac{L_{3}}{C_{4}} + j (ω L_{3} R_{4} - \frac{R_{3}}{ω C_{4}})$
$\Rightarrow R_{1} R_{2} = R_{3} R_{4} + \frac{L_{3}}{C_{4}}$ and $ω L_{3} R_{4} = \frac{R_{3}}{ω C_{4}}$
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.
$z = \frac{R_{4} (\frac{1}{j ω C_{4}})}{R_{4} + \frac{1}{j ω C_{4}}} = \frac{R_{4}}{1 + j ω R_{4} C_{4}}$
At bridge balance condition,
$R_{1} R_{2} = (R_{3} + j ω L_{3}) (\frac{R_{4}}{1 + j ω R_{4} C_{4}})$
⇒ R₁R₂ + jωR₁R₂R₄C₄ = R₃R₄ + jωL₃R₄
$\Rightarrow R_{4} = \frac{R_{1} R_{2}}{R_{3}} a n d C_{4} = \frac{L_{3}}{R_{1} R_{2}}$
Now it is easy to obtain bridge balance condition. So, the z is a parallel combination of R and C.
Question 8/10
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Consider the unbalanced Wheatstone bridge shown in figure.
The internal resistance of the galvanometer is 1 ohm. The galvanometer current is ___________ (in mA)

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A
4.9-5

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B
4.94-51

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C
4.97-56

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D
4.99-52

Solutions

V_th = V_a - V_b
$= 10 (\frac{100}{100 + 100}) - 10 (\frac{90}{110 + 90}) = 0.5 V$
R_th = R_ab = (110 || 90) + (100 || 100)
= 99.5Ω
$I_{g} = \frac{0.5}{100.5} = 4.975 m A$
Question 9/10
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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)

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A
86-87

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B
864-871

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C
867-876

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D
869-872

Solutions

Insulation resistance of the cable is,
$R = \frac{0.4343 t}{C l o g (\frac{V}{v})}$
$\Rightarrow R = \frac{0.4343 \times 60}{500 \times 10^{- 6} \log (\frac{400}{100})}$
⇒ R = 86.56 kΩ
Question 10/10
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An unknown voltage source E_u (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 E_u is

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A
0.9910 V

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B
0.9991 V

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C
1.0012 V

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D
1.0048 V

Solutions

By applying KVL,
-1.6 + 100 I_w + 500 I_w + 1000 (I_w + I_g) = 0
⇒ 1600 I_w + 1000 I_g = 1.6
⇒ 1600 I_w + 1000 (10 × 10^-6) = 1.6
⇒ 1600 I_w + 0.01 = 1.6
$\Rightarrow I_{w} = \frac{1.6 - 0.01}{1600} = 9.9375 \times 10^{- 4} A$
By applying KVL,
-E_u + I_g(100) + (I_w + I_g) (1000) = 0
⇒ E_u = 1000 I_w + 1100 I_g
= 0.99375 + 0.011 = 1.00475 V

Question 1/10

V0=13V

V0=14V

V0=15V

16V

Question 2/10

2-2

24-21

27-26

29-22

Question 3/10

0.19-0.21

0.194-0.211

0.197-0.216

0.199-0.212

Question 4/10

Question 5/10

0.003

0.002

0.004

0.001

Question 6/10

0.09-0.11

0.094-0.111

0.097-0.116

0.099-0.112

Question 7/10

an R and L in series

an R and C in series

an R and C in parallel

an R and L in parallel

Question 8/10

4.9-5

4.94-51

4.97-56

4.99-52

Question 9/10

86-87

864-871

867-876

869-872

Question 10/10

0.9910 V

0.9991 V

1.0012 V

1.0048 V

$V_{0} = \frac{1}{3} V$

$V_{0} = \frac{1}{4} V$

$V_{0} = \frac{1}{5} V$

$\frac{1}{6} V$