Question 1 |
The dependent current source shown in Figure
delivers 80 W | |
absorbs 80 W | |
delivers 40 W | |
absorbs 40 W |
Question 1 Explanation:
Applying KVL, 20-5 I-5\left(I+\frac{V_{1}}{5}\right)=0
20-10 I-20=0
\Rightarrow \quad I=0
\therefore Only dependent source acts.
\frac{V_{1}}{5}=4 \mathrm{A}
Power delivered =I^{2} R=16 \times 5=80 \mathrm{W}
20-10 I-20=0
\Rightarrow \quad I=0
\therefore Only dependent source acts.
\frac{V_{1}}{5}=4 \mathrm{A}
Power delivered =I^{2} R=16 \times 5=80 \mathrm{W}
Question 2 |
In figure, the switch was closed for a long time before opening at t = 0. the
voltage V_x at t = 0^{+} is
25 V | |
50 V | |
-50 V | |
0 V |
Question 2 Explanation:
When switch was closed circuit was in steady state,
i_{L}(O-) = 2.5 A
At, t = 0^{+}
\begin{aligned} \Rightarrow & & V=I R \\ & &=2.5 \times 20=50 \mathrm{V} \end{aligned}
\therefore \quad V_{x}=-50 \mathrm{V}
(Polarity of V_{x} is given reverse of n)
i_{L}(O-) = 2.5 A
At, t = 0^{+}
\begin{aligned} \Rightarrow & & V=I R \\ & &=2.5 \times 20=50 \mathrm{V} \end{aligned}
\therefore \quad V_{x}=-50 \mathrm{V}
(Polarity of V_{x} is given reverse of n)
Question 3 |
Convolution of x(t+5) with impulse function \delta(t-7) is equal to
x(t - 12) | |
x(t + 12) | |
x(t - 2) | |
x(t + 2) |
Question 3 Explanation:
\begin{aligned} x(t+5) * \delta(t-7) &=x(t+5-7) \\ &=x(t-2) \end{aligned}
Question 4 |
Which of the following cannot be the Fourier series expansion of a periodic
signal?
x(t) = 2cos t + 3 cos 3t | |
x(t) = 2cos \pit + 7 cos t | |
x(t) = cos t + 0.5 | |
x(t) = 2cos 1.5\pit + sin 3.5 \pit |
Question 4 Explanation:
\begin{aligned} x(t) &=2 \cos \pi t+7 \cos t \\ T_{1} &=\frac{2 \pi}{(0)}=2 \\ T_{2} &=\frac{2 \pi}{1}=2 \pi \\ \frac{T_{1}}{T_{2}} &=\frac{1}{\pi}=\text { irrational } \end{aligned}
\therefore x(t) is not periodic and does not satisfy Dirichlet condition.
\therefore x(t) is not periodic and does not satisfy Dirichlet condition.
Question 5 |
In Figure shown below, a silicon is carrying a constant current of 1 mA. When the
temperature of the diode is 20^{\circ}C, V_{D} is found to be 700 mV. If the temperature
rises to 40^{\circ}C, V_{D} becomes approximately equal to
740 mV | |
660 mV | |
680 mV | |
700 mV |
Question 5 Explanation:
For either Si or Ge\frac{d V}{d T} \simeq-2 \mathrm{mV} /^{\circ} \mathrm{C}
in order to maintain a constant value of I.
\begin{aligned} T_{2}-T_{1}&=40-20=20^{\circ} \mathrm{C} \\ -2 \times 20 \mathrm{mV}&=40 \mathrm{mV} \\ \text { Therefore, } \quad V_{D}&=700-40=660 \mathrm{mV} \end{aligned}
in order to maintain a constant value of I.
\begin{aligned} T_{2}-T_{1}&=40-20=20^{\circ} \mathrm{C} \\ -2 \times 20 \mathrm{mV}&=40 \mathrm{mV} \\ \text { Therefore, } \quad V_{D}&=700-40=660 \mathrm{mV} \end{aligned}
There are 5 questions to complete.