# 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}

There are 5 questions to complete.