Measurement of Resistance and Potentiometers

Question 1
An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the magnitude of V_o be maximum?
A
\sqrt{(1+x)}
B
(1+x)
C
1/\sqrt{(1+x)}
D
\sqrt{(1-x)}
GATE EE 2015-SET-1   Electrical and Electronic Measurements
Question 1 Explanation: 


\begin{aligned} V_0 &= \left [ \frac{R(1+x)}{PR+R(1+x)}-\frac{R}{R+PR} \right ]E\\ V_0 &=\left [ \frac{R(1+x)}{R+(1+x+P)}-\frac{R}{R+(1+P)} \right ]E \\ V_0&= \left [ \frac{1+x}{1+x+P}-\frac{1}{1+P} \right ]E\\ V_0&=\left [ \frac{(1+x)(1+P)-(1+x+P)}{(1+x+P)(1+P)} \right ]E \\ V_0&=\left [ \frac{1+x+Px+P-1-x-P}{1+P+x+xP+P+P^2} \right ]E \\ V_0&=\left [ \frac{Px}{1+2P+P^2+x+xP} \right ]E \end{aligned}
For maximum V_0,
\frac{dV_0}{dP}=0
\Rightarrow \;\frac{x}{1+x+P(P+2+x)}-\frac{xP(2P+x+2)}{1+x+P(2+P+x)^2}=0
\Rightarrow \; [1+x+P(2+P+x)]=P(2P+x+2)
\Rightarrow \; 1+x+2P+P^2+Px=2p^2+Px+2P
\Rightarrow \; 1+x=P^2
\Rightarrow \; P=\sqrt{1+x}
Question 2
Suppose that resistors R_1 \; and \; R_2 are connected in parallel to give an equivalent resistor R. If resistors R_1 \; and \; R_2 have tolerance of 1% each, the equivalent resistor R for resistors R_1 = 300 \Omega and R_2 = 200 \Omega will have tolerance of
A
0.50%
B
1%
C
1.20%
D
2%
GATE EE 2014-SET-2   Electrical and Electronic Measurements
Question 2 Explanation: 
Given,R_1=300\Omega , R_2=200\Omega
Equivalent resistance of parallel combination of R_1 and R_2

\begin{aligned} R &=\frac{R_1R_2}{R_1+R_2} =\frac{300 \times 200}{500}\\ &= 120\Omega =R_T\\ \text{Given tolerance of} \; R_1&=1\% \\ \therefore \; R_1 &= 300 \pm 3\\ R_1 &= 303 \Omega \; \text{or} 297\Omega \\ \text{Tolerance of} \; R_2&=1\% \\ \therefore \; R_2 &= 200 \pm 2\\ R_2 &= 202 \Omega \; \text{or} 198\Omega \end{aligned}
Equivalent resistance of the parallel combination of R_1 and R_2 in high range is:
R_H=\frac{303 \times 202}{303+202}=121.2\Omega
\therefore Tolerance for equivalent resistance R in high range of R_1 and R_2 is
\begin{aligned} \varepsilon _R &=\left ( \frac{R_H-R}{R} \right ) \times 100\% \\ &= \left ( \frac{121.2-120}{120} \right ) \times 100\% \\ &= \frac{1.2}{120} \times 100\%=1\% \end{aligned}
Note: In low range of R_1 and R_2,
R_L=\left ( \frac{297 \times 198}{495} \right )=118.8\Omega
\therefore Tolerance of equivalent resistance, R in low range is,
\begin{aligned} \varepsilon _L &=\left ( \frac{R_L-R}{R} \right ) \times 100\% \\ &= \left ( \frac{118.8-120}{120} \right ) \times 100\% \\ &= -1\% \end{aligned}
Question 3
A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as R_s=300\Omega. Other bridge resistances are R_1=R_2=R_3=300\Omega. The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage V_0 in mV is
A
56.02
B
40.83
C
29.85
D
10.02
GATE EE 2013   Electrical and Electronic Measurements
Question 3 Explanation: 


V_{ab}=V_a-V_b
Using voltage division rule,
V_{ab}=\left ( \frac{300}{300+300}-\frac{300}{300+303} \right ) \times 12
\;\;\;=29.85 mV
Question 4
R_1 \; and \; R_4 are the opposite arms of a Wheatstone bridge as are R_3 \; and \; R_2. The source voltage is applied across R_1 \; and \; R_3. Under balanced conditions which one of the following is true
A
R_1 =R_3R_4/R_2
B
R_1 =R_2R_3/R_4
C
R_1 =R_2R_4/R_3
D
R_1 =R_2+R_3+R_4
GATE EE 2006   Electrical and Electronic Measurements
Question 4 Explanation: 


Adjustment are made in various arms of the bridge so that the voltage across the detector is zero and hence no current slows through it, when no current flows through detector the bridge is said to be balanced.
Under condition of balance =R_1=R_2\frac{R_3}{R_4}
Question 5
The set-up in the figure is used to measure resistance R. The ammeter and voltmeter resistances are 0.01\Omega and 2000 \Omega, respectively. Their readings are 2A and 180V, respectively, giving a measured resistances of 90 \Omega. The percentage error in the measurement is
A
2.25%
B
2.35%
C
4.50%
D
4.71%
GATE EE 2005   Electrical and Electronic Measurements
Question 5 Explanation: 
Measured value of resistance =90\Omega
Resistance of voltmeter, R_v=2000 \Omega
Voltage across voltmeter, V=180
Current through voltmeter =\frac{V}{R_v} =\frac{180}{2000}=0.09A
Current through resistance,
(R)=I_R=2-I_v=2-0.09
\Rightarrow \;\;I_R=1.91A
True value of resistance,
\frac{V}{I_R}=\frac{180}{1.91}=94.24\Omega
\% erro=\frac{\text{measured value}-\text{True value}}{\text{True value}}=\frac{90-94.24}{94.24}\times 100=-4.5\%
Question 6
A dc potentiometer is designed to measure up to about 2 V with a slide wire of 300 mm. A standard cell of emf 1.18 V obtains balance at 600 mm. A test cell is seen to obtain balance at 680 mm. The emf of the test cell is
A
1.00 V
B
1.34 V
C
1.50 V
D
1.70 V
GATE EE 2004   Electrical and Electronic Measurements
Question 6 Explanation: 
\begin{aligned} E_1&= \text{standerd call of emf 1.18} \\ l_1&= 600mm\\ E_2&= \text{emf of the test cell}\\ i_2&= 680mm \end{aligned}
The voltage of any point along the slide wire is proportional to length of slide wire.
E\propto l
\frac{E_1}{E_2}=\frac{l_1}{l_2}
E_2=\frac{l_2}{l_1}E_1
\;\;\;=\frac{680}{600} \times 1.18=1.34V
There are 6 questions to complete.
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