# A.C. Bridges

 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
 A $30\Omega \pm 3\Omega$ B $30\Omega \pm 0.9\Omega$ C $3000\Omega \pm 90\Omega$ D $3000\Omega \pm 3\Omega$
GATE EE 2022   Electrical and Electronic Measurements
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}
 Question 2
Inductance is measured by
 A Schering bridge B Maxwell bridge C Kelvin bridge D Wien bridge
GATE EE 2021   Electrical and Electronic Measurements
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.
 Question 3
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 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 _________
 A 80.32 B 141.42 C 160.45 D 180.78
GATE EE 2014-SET-1   Electrical and Electronic Measurements
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)
$=V=V_{ab}=141.42$
 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
 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 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
 A Heaviside Campbell bridge B Schering bridge C De Sauty bridge D Wien bridge
GATE EE 2012   Electrical and Electronic Measurements
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 $\mu$F 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 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}
 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
 A low resistance B high resistance C low Q inductor D lossy capacitor
GATE EE 2011   Electrical and Electronic Measurements
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. $Q \lt 1$).
 Question 9
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 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
 A (260 + j0) $\omega$ B (0 + j200) $\omega$ C (260 - j200) $\omega$ D (260 + j200) $\omega$
GATE EE 2008   Electrical and Electronic Measurements
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.