A.C. Bridges

Question 1
In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance C_1 in microfarad is ______.
A
0.1
B
0.2
C
0.3
D
0.4
GATE EE 2014-SET-3   Electrical and Electronic Measurements
Question 1 Explanation: 


Redrawing the given bridge circuit, we have:
\begin{aligned} Z_1&= \frac{1}{j\omega C_1}\Omega \\ Z_2&=35 k\Omega Z_3 &= \frac{10^6}{j0.1\omega } \Omega \\ Z_4&=105 k\Omega \end{aligned}
At balance, current through galavanometer:
I_g=0
and |Z_1||Z_4|=|Z_2||Z_3|
\therefore \left ( \frac{1}{\omega C_1} \right ) \times (105K)=(35K)\left ( \frac{10^6}{0.1\omega } \right )
C_1=\frac{105 \times 0.1}{35 }\mu F
\;\;=0.3\mu F
Question 2
The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is _________
A
80.32
B
141.42
C
160.45
D
180.78
GATE EE 2014-SET-1   Electrical and Electronic Measurements
Question 2 Explanation: 
Given bridge is shown in figure

From the given bridge circuit, we see that product of opposite arm impedance are equal
\begin{aligned} i.e. \;\;& (-j) \times (-j)=(j)(j)\\ or\;\;\; & -1= -1 \\ \end{aligned}
Hence, the bridge is balanced, i.e. no current flows through the voltage source V.
Now, \begin{aligned} v_i(t)&= 100 \sin \omega t\\ &= \frac{100}{\sqrt{2}} \text{Volt(rms value)}\\ \end{aligned}
\therefore Net current supplied by the source is
I=\frac{100\sqrt{2}}{(0.5+Z_{eq})}A
Z_{eq}=0 (i.e. net impedance of whole bridge=0)
\therefore \;\; I=\frac{100\sqrt{2}}{0.5}=141.42A
\therefore \;\;\frac{I_2}{2}=70.71A
Current flow through each parallel path of the bridge circuit, therefore V=V_a-V_b
\;\;=\left ( \frac{I}{2} \times -j \right )-\left ( \frac{I}{2} \times j \right )
\;\;=-jI=-j141.42 volt (rms value)
\therefore Reading of the voltmeter in volt (rms)
=V=V_{ab}=141.42
Question 3
Three moving iron type voltmeters are connected as shown below. Voltmeter readings are V, V_1 \; and \;V_2 as indicated. The correct relation among the voltmeter readings is
A
V=\frac{V_{1}}{\sqrt{2}}+\frac{V_{2}}{\sqrt{2}}
B
V=V_{1}+V_{2}
C
V=V_{1}V_{2}
D
V=V_{2}-V_{1}
GATE EE 2013   Electrical and Electronic Measurements
Question 3 Explanation: 
\begin{aligned} V_1&= -|j1\Omega | \times I \\ V_2 &= |j2\Omega | \times I \\ V&= -j1\Omega | \times I +j2\Omega | \times I \\ &=V_2-V_1 \end{aligned}
Question 4
The bridge method commonly used for finding mutual inductance is
A
Heaviside Campbell bridge
B
Schering bridge
C
De Sauty bridge
D
Wien bridge
GATE EE 2012   Electrical and Electronic Measurements
Question 4 Explanation: 
Heaciside Campbell bridge method is commonly used for finding mutual inductance.
Question 5
A lossy capacitor C_x, rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor C_p in parallel with a resistor R_p. The value C_p is found to be 0.102 \muF and value of Rp = 1.25M\Omega. Then the power loss and \tan \delta of the lossy capacitor operating at the rated voltage, respectively, are
A
10 W and 0.0002
B
10 W and 0.0025
C
20 W and 0.025
D
20 W and 0.04
GATE EE 2011   Electrical and Electronic Measurements
Question 5 Explanation: 
\begin{aligned} C_P &=0.102\mu F \\ R_P &= 1.25M\Omega \\ \end{aligned}
\begin{aligned} \text{Power loss}&=\frac{V^2}{R_P} =\frac{(5 \times 10^3)^2}{1.25 \times 10^6}=\frac{25}{1.25}\\ &=20W \\ \tan\delta &=RIX=\frac{1}{\omega C_PR_P} \\ &=\frac{1}{2 \times \pi \times 50 \times 0.102 \times 10^{-6}\times 1.25 \times 10^6} \\ &= 0.025 \end{aligned}
Question 6
The bridge circuit shown in the figure below is used for the measurement of an unknown element Z_X. The bridge circuit is best suited when Z_X is a
A
low resistance
B
high resistance
C
low Q inductor
D
lossy capacitor
GATE EE 2011   Electrical and Electronic Measurements
Question 6 Explanation: 
The bridge is Maxwell bridge.
Element is an inductor.
Inductance =L_x effective resistance of the inductor =R_x
Q=\frac{\omega L_x}{R_x}=\omega C_1R_1 \;\;...(i)
The bridge is limited to measurement of low Q inductor (1 \lt Q \lt 10).
It is clear from equation (i) that the measurement of high Q coils demands a large value of resistance R_1, perhaps 10^5 or 10^6 \Omega . The resistance boxes of such high value are very expensive. Thus for value of Q \gt 10, the bridge is unsuitable.
The bridge is also unsuited for coils with a very low value of Q(i.e. Q \lt 1).
Question 7
The Maxwell's bridge shown in the figure is at balance. The parameters of the inductive coil are.
A
R=R_{2}R_{3}/R_{4},L=C_{4}R_{2}R_{3}
B
L=R_{2}R_{3}/R_{4},R=C_{4}R_{2}R_{3}
C
R=R_{4}/R_{2}R_{3},L=C_{4}R_{2}R_{3}
D
L=R_{4}/R_{2}R_{3},R=1/(C_{4}R_{2}R_{3})
GATE EE 2010   Electrical and Electronic Measurements
Question 7 Explanation: 


Z_1=R+j\omega L
Z_2=R_2
Z_3=R_3
Z_4=R_4||\left ( \frac{-j}{\omega C_4} \right )
\;\;=R_4||\left ( \frac{1}{\omega C_4} \right )
\;\;=\frac{R_4}{1+j\omega C_4R_4}
At balance,
Z_1Z_4=Z_2Z_3
(R+ j\omega L)\left ( \frac{R_4}{1+j\omega C_4R_4} \right )=R_2R_3
R_4(R+ j\omega L)=R_2R_3(1+j\omega C_4R_4)
RR_4+j\omega L R_4=R_2R_3+j\omega R_2R_3R_4C_4
Equating real and imaginary terms
RR_4=R_2R_3
R=\frac{R_2R_3}{R_4}
\omega LR_4=\omega R_2R_3R_4C_4
L=R_2R_3C_4
Question 8
The ac bridge shown in the figure is used to measure the impedance Z .

If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be
A
(260 + j0) \omega
B
(0 + j200) \omega
C
(260 - j200) \omega
D
(260 + j200) \omega
GATE EE 2008   Electrical and Electronic Measurements
Question 8 Explanation: 


Z_{AB}=500\Omega
Z_{CD}=Z
Z_{BC}=R_{BC}+\frac{1}{j\omega C}
\;\;=300-\frac{j}{2 \pi \times 2 \times 10^3 \times 0.398 \times 10^{-6}}
Z_{BC}=300-j200\Omega
Z_{AD}=R_{AD}+j\omega L
Z_{AD}=300+j2\pi\times 2 \times 10^3 \times 15.91 \times 10^{-3}
\;\;=300+j200\Omega
At balance,
Z_{AB} \times Z_{CD}=Z_{BC} \times Z_{AD}
Z_{CD}=Z
\;\;=\frac{Z_{BC} \times Z_{AD}}{Z_{AB}}
\Rightarrow \;Z=\frac{(300-j200)\times (300+j200)}{500}
Z=(260+j0)\Omega
Question 9
A bridge circuit is shown in the figure below. Which one of the sequence given below is most suitable for balancing the bridge ?
A
First adjust R_4, and then adjust R_1
B
First adjust R_2, and then adjust R_3
C
First adjust R_2, and then adjust R_4
D
First adjust R_4, and then adjust R_2
GATE EE 2007   Electrical and Electronic Measurements
Question 9 Explanation: 
x_1=\omega L_1 and x_4=\frac{1}{\omega C_4}
z_1=R_1+jx_1=R_1+j\omega L_1
z_2=R_2
z_3=R_3
z_4=R_4-jx_4=R_4-\frac{j}{\omega C_4}
Under balanced condition
z_1z_4=z_2z_3
(R_1+j\omega L_1)(R_4-jx_4)=R_2R_3
\left ( R_1R_4+\frac{L_1}{C_4} \right )+j\left ( \omega L_1R_4-\frac{R_1}{\omega C_4} \right )=R_2R_3
Equating real and imaginary terms, we obtain
R_1R_4+\frac{L_1}{C_4} =R_2R_3
\omega L_1R_4-\frac{R_1}{\omega C_4}=0
Solving above equations, we get
L_1=\frac{R_2R_3C_4}{1+\omega ^2C_4^2R_4^2}
and R_2=\frac{\omega ^2R_2R_3R_4C_4^2}{1+\omega ^2C_4^2R_4^2}
Q-factor of the coil Q=\frac{\omega L_1}{R_1}=\frac{1}{\omega C_4R_4}
Therefore, L_1=\frac{R_2R_3C_4}{1+\left ( \frac{1}{Q} \right )^2}\;\;...(i)
R_1=\frac{\omega ^2R_2R_3R_4C_4^2}{1+\left ( \frac{1}{Q} \right )^2}\;\;...(ii)
For a value of a greater than 10, the term (1/Q)^2 will be smaller than 1/1000 and can be neglected. Therefore equation (i) and (ii) reduces to
L_1=R_2R_3C_4 \;\;...(iii)
R_1=\omega ^2R_2R_3R_4C_4^2\;\;...(iv)
R_4 appears only in equation (iv) and R_2 appears in both equation (iii) and (iv).
So first R_2 is adjusted and then R_4 is adjusted.
Question 10
The items in Group-I represent the various types of measurements to be made with a reasonable accuracy using a suitable bridge. The items in Group-II represent the various bridges available for this purpose. Select the correct choice of the item in Group-II for the corresponding item in Group-I from the following
A
P=2, Q=3, R=6, S=5
B
P=2, Q=6, R=4, S=5
C
P=2, Q=3, R=5, S=4
D
P=1, Q=3, R=2, S=6
GATE EE 2003   Electrical and Electronic Measurements
Question 10 Explanation: 
Wheat stone bridge is used for measurement of medium resistance.
Kelvin double bridge is used for meserement of low resistance.
Schering bridge is used for meserement of low value of capacitances.
Wein's bridge is used for meserement of the frequency
Hay's bridge is used for meserement of inductance of a coil with a large time-constant.
Carey-foster bridge is used for comparision of resistances which are nearly equal.
There are 10 questions to complete.
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