1
GATE EE 2002
MCQ (Single Correct Answer)
+1
-0.3
Given a vector field $${\overrightarrow F ,}$$ the divergence theorem states that
A
$$\int\limits_s {\overrightarrow F .d\overrightarrow s = \int\limits_v \nabla .\overrightarrow F \,dv} $$
B
$$\int\limits_s {\overrightarrow F .d\overrightarrow s = \int\limits_v \nabla \times \overrightarrow F \,dv} $$
C
$$\int\limits_s {\overrightarrow F \times d\overrightarrow s = \int\limits_v \nabla .\overrightarrow F \,dv} $$
D
$$\int\limits_s {\overrightarrow F \times d\overrightarrow s = \int\limits_v \nabla \times \overrightarrow F \,dv} $$
2
GATE EE 2002
MCQ (Single Correct Answer)
+1
-0.3
The determinant of the matrix $$\left[ {\matrix{ 1 & 0 & 0 & 0 \cr {100} & 1 & 0 & 0 \cr {100} & {200} & 1 & 0 \cr {100} & {200} & {300} & 1 \cr } } \right]$$ is
A
$$100$$
B
$$200$$
C
$$1$$
D
$$300$$
3
GATE EE 2002
Subjective
+5
-0
In fig, the ideal switch $$S$$ is switched on and off with a switching frequency $$f = 10 kHz.$$ The switching time period is $$\,T = {t_{ON}} + t{}_{off} = 100\,\,\mu s.$$ The circuit is operated in steady state at the boundary of continuous and discontinuous conduction, so that the inductor current i is as shown in Fig.P20. Find

(a) The on-time $${t_{ON}}$$ of the switch
(b) The value of the peak current $${{\rm I}_p}$$

GATE EE 2002 Power Electronics - Choppers and Commutation Techniques Question 19 English 1 GATE EE 2002 Power Electronics - Choppers and Commutation Techniques Question 19 English 2
4
GATE EE 2002
MCQ (Single Correct Answer)
+2
-0.6
In the chopper circuit shown in fig the input $$dc$$ voltage has a constant value $${{V_s}}$$. the output voltage $${{V_0}}$$ is assumed ripple free. The switch $$S$$ is operated with a switching time period $$T$$ and a duty ratio $$D.$$ What is the value of $$D$$ at the boundary of continuous and discontinuous conduction of the inductor current $${i_L}$$? GATE EE 2002 Power Electronics - Choppers and Commutation Techniques Question 39 English
A
$$D = 1 - {{{V_s}} \over {{V_0}}}$$
B
$$D = {{2L} \over {RT}}$$
C
$$D = 1 - {{2L} \over {RT}}$$
D
$$D = {{RT} \over L}$$