1
GATE EE 2010
MCQ (Single Correct Answer)
+2
-0.6
The transistor circuit shown uses a silicon transistor with $${V_{BE}} = 0.7V,{{\rm I}_C} \approx {{\rm I}_E}$$ and a $$DC$$ current gain of $$100.$$ The value of $${V_0}$$ is GATE EE 2010 Analog Electronics - Bjt and Mosfet Biasing Question 15 English
A
$$4.65$$ $$V$$
B
$$5$$ $$V$$
C
$$6.3$$ $$V$$
D
$$7.32$$ $$V$$
2
GATE EE 2010
MCQ (Single Correct Answer)
+1
-0.3
Given that the op-amp is ideal, the $$O/P$$ voltage $${V_0}$$ is GATE EE 2010 Analog Electronics - Operational Amplifier Question 61 English
A
$$4V$$
B
$$6V$$
C
$$7.5V$$
D
$$12.12V$$
3
GATE EE 2010
MCQ (Single Correct Answer)
+1
-0.3
As shown in the figure, a negative feedback system has an amplifier of gain $$100$$ with $$ \pm 10\% $$ tolerance in the forward path, and an attenuator of value $$9/100$$ in the feedback path. The overall system gain is approximately. GATE EE 2010 Analog Electronics - Feedback Amplifiers and Oscillator Circuits Question 10 English
A
$$10 \pm 1\% $$
B
$$10 \pm 2\% $$
C
$$10 \pm 5\% $$
D
$$10 \pm 10\% $$
4
GATE EE 2010
MCQ (Single Correct Answer)
+2
-0.6
The frequency response of $$G\left( s \right) = 1/\left[ {s\left( {s + 1} \right)\left( {s + 2} \right)} \right]$$ plotted in the complex $$\,G\left( {j\omega } \right)$$ plane $$\left( {for\,\,0 < \omega < \infty } \right)$$ is
A
GATE EE 2010 Control Systems - Polar Nyquist and Bode Plot Question 25 English Option 1
B
GATE EE 2010 Control Systems - Polar Nyquist and Bode Plot Question 25 English Option 2
C
GATE EE 2010 Control Systems - Polar Nyquist and Bode Plot Question 25 English Option 3
D
GATE EE 2010 Control Systems - Polar Nyquist and Bode Plot Question 25 English Option 4
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