# 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.