Q2GATE 2015MCQ

An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the magnitude of $V_o$ be maximum?

📖 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 $V_0,$ $\frac{dV_0}{dP}=0$ $\Rightarrow \;\frac{x}{1+x+P(P+2+x)}-\frac{xP(2P+x+2)}{1+x+P(2+P+x)^2}=0$ $\Rightarrow \; [1+x+P(2+P+x)]=P(2P+x+2)$ $\Rightarrow \; 1+x+2P+P^{2+}Px=2p^{2+}Px+2P$ $\Rightarrow \; 1+x=P^2$ $\Rightarrow \; P=\sqrt{1+x}$

Q3GATE 2014MCQ

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

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📖 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}

Q4GATE 2013MCQ

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

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📖 Explanation

$V_{ab}=V_a-V_b$ Using voltage division rule, $V_{ab}=\left ( \frac{300}{300+300}-\frac{300}{300+303} \right ) \times 12$ $\;\;\;=29.85 mV$

Q5GATE 2006MCQ

$R_1 \; and \; R_4$ 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

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📖 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 $=R_1=R_2\frac{R_3}{R_4}$

Q6GATE 2005MCQ

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

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📖 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, $\frac{V}{I_R}=\frac{180}{1.91}=94.24\Omega$ $\% erro=\frac{\text{measured value}-\text{True value}}{\text{True value}}=\frac{90-94.24}{94.24}\times 100=-4.5\%$

Q7GATE 2004MCQ

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

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📖 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. $E\propto l$ $\frac{E_1}{E_2}=\frac{l_1}{l_2}$ $E_2=\frac{l_2}{l_1}E_1$ $\;\;\;=\frac{680}{600} \times 1.18=1.34V$

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