Question 1 |
The temperature of the coolant oil bath for a transformer is monitored using the circuit
shown. It contains a thermistor with a temperature-dependent resistance,
R_{thermistor} = 2(1 + \alpha T) k\Omega. Where T is the temperature in ^{\circ}C. The temperature coefficient
\alpha, is -(4 \pm 0.25) \%/^{\circ}C. Circuit parameters: R_1 = 1 k\Omega , R_2 = 1.3 k\Omega , R_3 = 2.6 k\Omega. The
error in the output signal (in V. rounded off lo 2 decimal places) at 150^{\circ}C is ________.
0.01 | |
0.08 | |
0.04 | |
0.06 |
Question 1 Explanation:
As per GATE official answer key MTA (Marks to ALL)
\begin{aligned} \text{Given data,}\\ R_{thermistor}&=R_{th}=2(1+\alpha T)K\Omega \\ \alpha &=-(4+0.25)\%/^{\circ}C=-(0.04\pm 0.0025)^{\circ}C \\ \alpha _{max}&=-0.0424/^{\circ}C, \; \; \alpha _{min}=-0.375/^{\circ}C\\ \text{Temperature, } T&=150^{\circ}C \\ R_{1}&=1 K\Omega ,\; R_{2}=1.3 K\Omega , \; R_{3}=2.6 K\Omega\\ \text{Considering, }\alpha &=-0.04 \\ R_{th}&=2[1-0.04\times 150]=-10 K\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k+10k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.5 V \\ \text{Case-1:} \\ \text{Considering, }\alpha _{max}&=-0.0425/^{\circ}C \\ R_{thmax}&=2[1+(-0.0425)\times 150] =-10.75\, k\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k-10.75k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.46 V \\ \text{Case-2:} \\ \text{Considering, }\alpha _{min}&=-0.0375/^{\circ}C \\ R_{thmin}&=2[1+(-0.0375)\times 150] =-9.25\, k\Omega \\ V_{0} &=3\times \frac{1k}{1k-9.25k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.54 V \\ \text{Output voltage, }\\ V_0&=0.5\pm0.04 \; \Rightarrow \; \text{Error}=0.04 \end{aligned}
\begin{aligned} \text{Given data,}\\ R_{thermistor}&=R_{th}=2(1+\alpha T)K\Omega \\ \alpha &=-(4+0.25)\%/^{\circ}C=-(0.04\pm 0.0025)^{\circ}C \\ \alpha _{max}&=-0.0424/^{\circ}C, \; \; \alpha _{min}=-0.375/^{\circ}C\\ \text{Temperature, } T&=150^{\circ}C \\ R_{1}&=1 K\Omega ,\; R_{2}=1.3 K\Omega , \; R_{3}=2.6 K\Omega\\ \text{Considering, }\alpha &=-0.04 \\ R_{th}&=2[1-0.04\times 150]=-10 K\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k+10k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.5 V \\ \text{Case-1:} \\ \text{Considering, }\alpha _{max}&=-0.0425/^{\circ}C \\ R_{thmax}&=2[1+(-0.0425)\times 150] =-10.75\, k\Omega \\ V_{0}&=V_{1}\times \frac{R_{1}}{R_{1}+R_{th}}\left [ 1+\frac{R_{2}}{R_{3}} \right ]\\ &=3\times \frac{1k}{1k-10.75k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.46 V \\ \text{Case-2:} \\ \text{Considering, }\alpha _{min}&=-0.0375/^{\circ}C \\ R_{thmin}&=2[1+(-0.0375)\times 150] =-9.25\, k\Omega \\ V_{0} &=3\times \frac{1k}{1k-9.25k}\left [ 1+\frac{1.3k}{2.6k} \right ] \\ &=-0.54 V \\ \text{Output voltage, }\\ V_0&=0.5\pm0.04 \; \Rightarrow \; \text{Error}=0.04 \end{aligned}
Question 2 |
A common-source amplifier with a drain resistance, R_D = 4.7 k\Omega, is powered using a
10 V power supply. Assuming that the transconductance, g_m, \;is\; 520 \mu A/V, the voltage
gain of the amplifier is closest to:
-2.44 | |
-1.22 | |
1.22 | |
2.44 |
Question 2 Explanation:
Given data:
R_{D}=4.7 K\Omega ,\; g_{m}=520 \mu A/V
Voltage gain of CS amplifier
=-g_{m}R_{D}=-520\, \mu A/V\, \times \, 4.7\, k\Omega =-2.44
R_{D}=4.7 K\Omega ,\; g_{m}=520 \mu A/V
Voltage gain of CS amplifier
=-g_{m}R_{D}=-520\, \mu A/V\, \times \, 4.7\, k\Omega =-2.44
Question 3 |
In the circuit below, the operational amplifier is ideal. If V_1=10 mV and V_2=50 mV, the output voltage (V_{out}) is
100 mV | |
400 mV | |
500 mV | |
600 mV |
Question 3 Explanation:
V_o=\frac{R_2}{R_1}(V_2-V_1)
\;\;=\frac{100k}{10k}(50\; mV-10\; mV)
\;\;=10(40\; mV)=400\;mV
Question 4 |
The op-amp shown in the figure is ideal. The input impedance \frac{v_{in}}{i_{in}} is given by
Z\frac{R_{1}}{R_{2}} | |
-Z\frac{R_{1}}{R_{2}} | |
Z | |
Z\frac{R_{1}}{R_{1}+R_{2}} |
Question 4 Explanation:
According to virtual ground,
\begin{aligned} V_A&=V_B=V_{in} \\ \text{At node A,} \\ \frac{V_{in}-V_o}{Z}&=i_{in}\;\; ...(i) \\ V_{in}&=\left [ \frac{1}{R_2}+\frac{1}{R_1} \right ]=\frac{V_o}{R_1} \\ V_{in}\left [ \frac{R_1+R_2}{R_1R_2} \right ] \times R_1&=V_o \\ V_o&=V_{in} \times \left [ \frac{R_1+R_2}{R_2} \right ] \;\;...(ii)\\ \text{Equation (ii)}& \text{ in euation (i), }\\ \frac{V_{in}-V_o}{Z}&=i_{in} \\ \frac{V_{in}-V_{in}\left [ \frac{R_1+R_2}{R_2} \right ]}{Z}&=i_{in} \\ \frac{V_{in}}{i_{in}}\left [ 1-\frac{R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}\left [\frac{R_2-R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}&=-Z\cdot \frac{R_2}{R_1} \end{aligned}
\begin{aligned} V_A&=V_B=V_{in} \\ \text{At node A,} \\ \frac{V_{in}-V_o}{Z}&=i_{in}\;\; ...(i) \\ V_{in}&=\left [ \frac{1}{R_2}+\frac{1}{R_1} \right ]=\frac{V_o}{R_1} \\ V_{in}\left [ \frac{R_1+R_2}{R_1R_2} \right ] \times R_1&=V_o \\ V_o&=V_{in} \times \left [ \frac{R_1+R_2}{R_2} \right ] \;\;...(ii)\\ \text{Equation (ii)}& \text{ in euation (i), }\\ \frac{V_{in}-V_o}{Z}&=i_{in} \\ \frac{V_{in}-V_{in}\left [ \frac{R_1+R_2}{R_2} \right ]}{Z}&=i_{in} \\ \frac{V_{in}}{i_{in}}\left [ 1-\frac{R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}\left [\frac{R_2-R_1+R_2}{R_2} \right ]&=Z \\ \frac{V_{in}}{i_{in}}&=-Z\cdot \frac{R_2}{R_1} \end{aligned}
Question 5 |
For the circuit shown below, assume that the OPAMP is ideal.
Which one of the following is TRUE?
Which one of the following is TRUE?
v_o=v_s | |
v_o=1.5v_s | |
v_o=2.5v_s | |
v_o=5v_s |
Question 5 Explanation:
V_A=\frac{V_s}{2}
I_3=\frac{V_A}{R}=\frac{V_s}{2R}
V_B-V_A=I_3R
V_B=V_A+I_3R=\frac{V_s}{2}+\frac{V_s}{2}=V_s
I_2=\frac{V_B}{R}=\frac{V_s}{R}
I_1=I_2+I_3
\;\;=\frac{V_s}{R}+\frac{V_s}{2R}=\frac{V_s}{R}[1.5]
V_0-V_B=I_1R
\Rightarrow V_0=V_B+\frac{V_s}{R}[1.5]R
\;\;\;=V_s+1.5V_s=2.5V_s
Question 6 |
The approximate transfer characteristic for the circuit shown below with an ideal operational
amplifier and diode will be
A | |
B | |
C | |
D |
Question 6 Explanation:
V_{in} \gt 0\;\; V_A=+V_{55}, D on, V_o=V_{in}
V_{in} \lt 0\;\; V_A=-V_{55}, D off, V_o=0
Question 7 |
For the circuit shown below, taking the opamp as ideal, the output voltage V_{out} in terms of the input voltages V_{1},V_{2} \; and \; V_{3} is
1.8V_{1}+7.2V_{2} -V_{3} | |
2V_{1}+8V_{2} -9V_{3} | |
7.2V_{1}+1.8V_{2} -V_{3} | |
8V_{1}+2V_{2} -9V_{3} |
Question 7 Explanation:
V_A=\left ( \frac{4}{5}V_1+\frac{1}{5}V_2 \right )
V_{out}=-9V_3+10V_A
\;\;=-9V_3+8V_1+2V_2
Question 8 |
The circuit shown below is an example of a
low pass filter. | |
band pass filter | |
high pass filter. | |
notch filter. |
Question 8 Explanation:
\frac{V_{out}}{V_{in}}=-\left [\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right ]
\frac{V_{out}}{V_{in}}=-\left [ \frac{R_2}{R_1(R_2j\omega C+1)} \right ]
So the system is a low pass filter.
Question 9 |
The saturation voltage of the ideal op-amp shown below is \pm10 V. The output voltage v_0 of the following circuit in the steady-state is
square wave of period 0.55 ms. | |
triangular wave of period 0.55 ms | |
square wave of period 0.25 ms. | |
triangular wave of period 0.25 ms. |
Question 9 Explanation:
The given circuit in a astable multivibrator, so output will be a periodic square wave and from the circuit \beta =0.5.
So, time period will be 2\tau \ln\left ( \frac{1+\beta }{1-\beta } \right )
T=2 \times RC \ln \left ( \frac{1+0.5}{1-0.5} \right )=0.55ms
So, time period will be 2\tau \ln\left ( \frac{1+\beta }{1-\beta } \right )
T=2 \times RC \ln \left ( \frac{1+0.5}{1-0.5} \right )=0.55ms
Question 10 |
The filters F1 and F2 having characteristics as shown in Figures (a) and (b) are connected as shown in Figure (c).
The cut-off frequencies of F1 and F2 are f_1 \; and \; f_2 respectively. If f_1 \lt f_2 , the resultant circuit exhibits the characteristic of a
The cut-off frequencies of F1 and F2 are f_1 \; and \; f_2 respectively. If f_1 \lt f_2 , the resultant circuit exhibits the characteristic of a
Band-pass filter | |
Band-stop filter | |
All pass filter | |
High-Q filter |
Question 10 Explanation:
To check the type of system:
we apply a delta function at input V_i=\delta (t)
V(f)=1\forall f
So, voltage at A will be same as voltage at output V_o. V_A will be equal to voltage due to F_1 + voltage due to F_2. Since, f_1 \lt f_2, so V_A/V_i will be
so, the system works as a band stop filter.
we apply a delta function at input V_i=\delta (t)
V(f)=1\forall f
So, voltage at A will be same as voltage at output V_o. V_A will be equal to voltage due to F_1 + voltage due to F_2. Since, f_1 \lt f_2, so V_A/V_i will be
so, the system works as a band stop filter.
There are 10 questions to complete.
