Question 1 |
In the figure given below the value of R is
2.5\Omega | |
5.0\Omega | |
7.5\Omega | |
10.0\Omega |
Question 1 Explanation:
The resultant (R) when viewed from voltage source =\frac{100}{8}=12.5
R+10||10=12.5\Omega
\therefore \;\; R=12.5-10||10 =12.5-5=7.5\Omega
R+10||10=12.5\Omega
\therefore \;\; R=12.5-10||10 =12.5-5=7.5\Omega
Question 2 |
The RMS value of the voltage u(t)= 3 + 4 cos (3t) is
\sqrt{17} V | |
5V | |
7V | |
(3+2\sqrt{2})V |
Question 2 Explanation:
R.M.S. value of d.c voltage =V_{dc}^{(rms)}=3V
R.M.S. value of a.c. voltage =V_{ac}^{(rms)}=\left ( \frac{4}{\sqrt{2}} \right ) V
Therefore, R.M.S. value of the voltage
\sqrt{3^2+\left ( \frac{4}{\sqrt{2}} \right )^2}
=\sqrt{9+8}=\sqrt{17}V
R.M.S. value of a.c. voltage =V_{ac}^{(rms)}=\left ( \frac{4}{\sqrt{2}} \right ) V
Therefore, R.M.S. value of the voltage
\sqrt{3^2+\left ( \frac{4}{\sqrt{2}} \right )^2}
=\sqrt{9+8}=\sqrt{17}V
Question 3 |
For the two port network shown in the figure, the Z-matrix is given by
\begin{bmatrix} Z_{1} &Z_{1}+Z_{2} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{2} \\ Z_{2} & Z_{1}+Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1} & Z_{1}+Z_{2} \end{bmatrix} |
Question 3 Explanation:
v_1=(i_1+i_2)Z_1 =Z_1i_1+Z_1i_2
v_2=Z_2i_1+Z_1(i_1+i_2) =Z_1i_1+(Z_1+Z_2)i_2
From above, we can write,
\text{Z-matrix}=\begin{bmatrix} Z_1 &Z_1 \\ Z_1 & Z_1+Z_2 \end{bmatrix}
Question 4 |
In the figure given, the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is
20V | |
10V | |
5V | |
0V |
Question 4 Explanation:
At (t\rightarrow 0^+). The capacitor act as short-circuit. At (t\rightarrow \infty ), the capacitor will become open circuit.
Therefore, voltage across capacitor =\frac{20}{10+10}\times 10=10V
Therefore, voltage across capacitor =\frac{20}{10+10}\times 10=10V
Question 5 |
If \vec{E} is the electric intensity, \triangledown \cdot (\triangledown \times \vec{E}) is equal to
\vec{E} | |
|\vec{E}| | |
null vector | |
Zero |
Question 5 Explanation:
Divergence of a curl field is always zero.
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
There are 5 questions to complete.