Question 1 |
A balanced Wheatstone bridge ABCD has the following arm resistances:
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
30\Omega \pm 3\Omega | |
30\Omega \pm 0.9\Omega | |
3000\Omega \pm 90\Omega | |
3000\Omega \pm 3\Omega |
Question 1 Explanation:
The condition for balanced bridge
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
Question 2 |
Inductance is measured by
Schering bridge | |
Maxwell bridge | |
Kelvin bridge | |
Wien bridge |
Question 2 Explanation:
Maxwell's bridge is used for measurement of inductance.
Wein's bridge is used for measurement of frequency
Kelvin's bridge is used for measurement of low value of resistance.
Schering bridge is used for measurement of capacitance, dilectric loss and permittivity etc.
Wein's bridge is used for measurement of frequency
Kelvin's bridge is used for measurement of low value of resistance.
Schering bridge is used for measurement of capacitance, dilectric loss and permittivity etc.
Question 3 |
In the bridge circuit shown, the capacitors are loss free. At balance, the value of
capacitance C_1 in microfarad is ______.


0.1 | |
0.2 | |
0.3 | |
0.4 |
Question 3 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 4 |
The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is
_________


80.32 | |
141.42 | |
160.45 | |
180.78 |
Question 4 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

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 5 |
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


V=\frac{V_{1}}{\sqrt{2}}+\frac{V_{2}}{\sqrt{2}} | |
V=V_{1}+V_{2} | |
V=V_{1}V_{2} | |
V=V_{2}-V_{1} |
Question 5 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 6 |
The bridge method commonly used for finding mutual inductance is
Heaviside Campbell bridge | |
Schering bridge | |
De Sauty bridge | |
Wien bridge |
Question 6 Explanation:
Heaciside Campbell bridge method is commonly used for finding mutual inductance.
Question 7 |
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
10 W and 0.0002 | |
10 W and 0.0025 | |
20 W and 0.025 | |
20 W and 0.04 |
Question 7 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}
\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 8 |
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


low resistance | |
high resistance | |
low Q inductor | |
lossy capacitor |
Question 8 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).
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 9 |
The Maxwell's bridge shown in the figure is at balance. The parameters of the
inductive coil are.

R=R_{2}R_{3}/R_{4},L=C_{4}R_{2}R_{3} | |
L=R_{2}R_{3}/R_{4},R=C_{4}R_{2}R_{3} | |
R=R_{4}/R_{2}R_{3},L=C_{4}R_{2}R_{3} | |
L=R_{4}/R_{2}R_{3},R=1/(C_{4}R_{2}R_{3}) |
Question 9 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 10 |
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

If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be
(260 + j0) \omega | |
(0 + j200) \omega | |
(260 - j200) \omega | |
(260 + j200) \omega |
Question 10 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
There are 10 questions to complete.