Question 1 |
A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across
its terminals from -5 V to +5 V. The effective thermal voltage, V_T, for the diode is
measured to be (29\pm2) mV. The resolution of the voltage source in the measurement
range is 1 mV. The percentage uncertainty (rounded off to 2 decimal plates) in the
measured current at a bias voltage of 0.02 V is _______.
5.87 | |
2.35 | |
11.5 | |
9.2 |
Question 2 |
Two resistors with nominal resistance values R_1 \; and\; R_2 have additive uncertainties
\Delta R_1 and \Delta R_2, respectively. When these resistances are connected in parallel, the standard deviation of the error in the equivalent resistance R is
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_1}\Delta R_1 \right \}^2+\left \{ \frac{\partial R}{\partial R_2}\Delta R_2 \right \}^2} | |
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_2}\Delta R_1 \right \}^2+\left \{ \frac{\partial R}{\partial R_1}\Delta R_2 \right \}^2} | |
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_1}\right \}^2 \Delta R_2 +\left \{ \frac{\partial R}{\partial R_2} \right \}^2 \Delta R_1} | |
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_1}\right \}^2 \Delta R_1 +\left \{ \frac{\partial R}{\partial R_2} \right \}^2 \Delta R_2} |
Question 2 Explanation:
\begin{aligned} \sigma _{res} &= \sqrt{\left ( \frac{\partial R}{\partial R_1} \right )^2 \sigma _1^2+\left ( \frac{\partial R}{\partial R_2} \right )^2 \sigma _2^2} \\ &= \sqrt{\left ( \frac{\partial R}{\partial R_1} \right )^2 \Delta R_1^2+\left ( \frac{\partial R}{\partial R_2} \right )^2 \Delta R_2^2} \end{aligned}
Question 3 |
The following measurements are obtained on a single phase load:
V = 220V\pm1%, I = 5.0A\pm1% and W=555W\pm 2%.
If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.
V = 220V\pm1%, I = 5.0A\pm1% and W=555W\pm 2%.
If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.
20% | |
40% | |
4% | |
0.40% |
Question 3 Explanation:
\begin{aligned} V &= 220 \pm 1\% \\ I&= 5 \pm 1 \%\\ W &=555 \pm 2 \% \\ W&= VI \cos (\phi )\\ p.f. &=\cos (\phi )=\frac{W}{VI} \\ &= \frac{555 \pm 2 \%}{(220 \pm 1\%)(5 \pm 1 \%)}\\ &= \frac{555}{220 \times 5} \pm 4\%\\ p.f.&= 0.5 \pm 4\% \end{aligned}
Question 4 |
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?
[123.50, 136.50] | |
[125.89, 134.12] | |
[117.00, 143.00] | |
[120.25, 139.75] |
Question 4 Explanation:
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
Question 5 |
The measurement system shown in the figure uses three sub-systems in cascade
whose gains are specified as G_1,G_2 \; and \; \frac{1}{G_3}. The relative small errors associated with
each respective subsystem G_1,G_2 \; and \; G_3 are \varepsilon _1,\varepsilon _2 \; and \; \varepsilon _3. The error associated
with the output is :
\varepsilon _{1}+ \varepsilon _{2} +1/ \varepsilon _{3} | |
\frac{\varepsilon _{1}\varepsilon _{2}} {\varepsilon _{3}} | |
\varepsilon _{1}+ \varepsilon _{2} -\varepsilon _{3} | |
\varepsilon _{1}+ \varepsilon _{2} +\varepsilon _{3} |
Question 5 Explanation:
\frac{dG_1}{G_1}=\varepsilon _1, \frac{dG_2}{G_2}=\varepsilon _2,and \frac{dG_3}{G_3}=\varepsilon _3
Output (y)_0=\frac{G_1G_3}{G_3}x
where, \;\; x=input
\ln y=\ln G_1+ \ln G_2- \ln G_3 +\ln x
Differentiating both side,
\frac{dy}{y}=\frac{dG_1}{G_1}+\frac{dG_2}{G_2}-\frac{dG_3}{G_3}+\frac{dx}{x}
No error is specified in input, so \frac{dx}{x}=0
\frac{dy}{y}==\varepsilon _1+\varepsilon _2-\varepsilon _3
Output (y)_0=\frac{G_1G_3}{G_3}x
where, \;\; x=input
\ln y=\ln G_1+ \ln G_2- \ln G_3 +\ln x
Differentiating both side,
\frac{dy}{y}=\frac{dG_1}{G_1}+\frac{dG_2}{G_2}-\frac{dG_3}{G_3}+\frac{dx}{x}
No error is specified in input, so \frac{dx}{x}=0
\frac{dy}{y}==\varepsilon _1+\varepsilon _2-\varepsilon _3
There are 5 questions to complete.