Question 1 |
The value of x for which the matrix A=\begin{bmatrix} 3 & 2&4 \\ 9& 7&13 \\ -6&-4 & -9+x \end{bmatrix}
has zero as an eigenvalue is ________
has zero as an eigenvalue is ________
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
A has an eigen value is zero
\begin{aligned} \therefore \quad|A|&=0 \\ \left| \begin{array}{lll} 3 & 2 & 4\\9 & 7 & 13 \\ -6 & -4 & -9+x\end{array}\right|&=0 \\ 3(-63+7 x+52)-2(-81+9 x+78)&+4(-36+42)=0 \\ 3(7 x-11)-2(9 x-3)+4(6)&=0 \\ 21 x-33-18 x+6+24&=0 \\ 3 x-3&=0 \\ x&=1 \end{aligned}
\begin{aligned} \therefore \quad|A|&=0 \\ \left| \begin{array}{lll} 3 & 2 & 4\\9 & 7 & 13 \\ -6 & -4 & -9+x\end{array}\right|&=0 \\ 3(-63+7 x+52)-2(-81+9 x+78)&+4(-36+42)=0 \\ 3(7 x-11)-2(9 x-3)+4(6)&=0 \\ 21 x-33-18 x+6+24&=0 \\ 3 x-3&=0 \\ x&=1 \end{aligned}
Question 2 |
Consider the complex valued function f(z)=2z^{3}+b|z|^{3} where z is a complex variable. The value of b for which the function f(z) is analytic is ________
1 | |
2 | |
3 | |
0 |
Question 2 Explanation:
f(z)=2 z^{3}+b_{1}|z|^{3}
Given that f(z) is analytic.
which is possible only when b=0
since \left|z^{3}\right| is differentiable at the origin but not analytic.
2 z^{3} is analytic everywhere
\begin{aligned} \therefore \quad f(z)&=2 z^{3}+b\left|z^{3}\right| \text{ is analytic}\\ \text{only when }\quad b&=0 \end{aligned}
Given that f(z) is analytic.
which is possible only when b=0
since \left|z^{3}\right| is differentiable at the origin but not analytic.
2 z^{3} is analytic everywhere
\begin{aligned} \therefore \quad f(z)&=2 z^{3}+b\left|z^{3}\right| \text{ is analytic}\\ \text{only when }\quad b&=0 \end{aligned}
Question 3 |
As x varies from -1 to +3, which one of the following describes the behaviour of the function f(z)=x^{3}-3x^{2}+1 ?
f(x) increases monotonically. | |
f(x) increases, then decreases and increases again. | |
f(x) decreases, then increases and decreases again. | |
f(x) increases and then decreases. |
Question 3 Explanation:
\begin{array}{l} f(x)=x^{3}-3 x^{2}+1 \\ f^{\prime}(x)=3 x^{2}-6 x \\ f(x)=0 \end{array}
\begin{aligned} 3 x^{2}-6 x &=0 \\ 3 x(x-2) &=0 \\ x &=0,2 \\ f^{\prime \prime}(x) &=6 x-6 \\ \text{At}\quad x &=0 \quad f^{\prime \prime}(0)=-6 \text { maxima } \\ x &=2 \quad f^{\prime \prime}(2)=6 \text { minima } \end{aligned}
\begin{aligned} 3 x^{2}-6 x &=0 \\ 3 x(x-2) &=0 \\ x &=0,2 \\ f^{\prime \prime}(x) &=6 x-6 \\ \text{At}\quad x &=0 \quad f^{\prime \prime}(0)=-6 \text { maxima } \\ x &=2 \quad f^{\prime \prime}(2)=6 \text { minima } \end{aligned}
Question 4 |
How many distinct values of x satisfy the equation sin(x)=x/2, where x is in radians?
1 | |
2 | |
3 | |
4 or more |
Question 4 Explanation:
Hence, 3 solutions.
Question 5 |
Consider the time-varying vector I=\hat{x}15cos(\omega t)+\hat{y}5sin(\omega t) in Cartesian coordinates, where \omega \gt 0 is a constant. When the vector magnitude |I| is at its minimum value, the angle \theta that I makes with the x axis (in degrees, such that 0 \leq \theta \leq 180) is ________
80 | |
90 | |
100 | |
110 |
Question 5 Explanation:
\begin{aligned} I &=\hat{x} 15 \cos \omega t+\hat{y} 5 \sin \omega t \\ \| I &=\sqrt{\left(15 \cos (t)^{2}+(5 \sin \omega t)^{2}\right.} \\ &=\sqrt{225 \cos ^{2} \omega t+25 \sin ^{2} \omega t} \\ &=\sqrt{25+200 \cos ^{2} \omega t} \end{aligned}
|I| is minimum when \cos ^{2}\omega t=0
text{or}\quad \theta=\omega t=90^{\circ}
|I| is minimum when \cos ^{2}\omega t=0
text{or}\quad \theta=\omega t=90^{\circ}
Question 6 |
In the circuit shown below, V_{S} is a constant voltage source and I_{L} is a constant current load.
The value of I_{L} that maximizes the power absorbed by the constant current load is
The value of I_{L} that maximizes the power absorbed by the constant current load is
\frac{V_{S}}{4R} | |
\frac{V_{S}}{2R} | |
\frac{V_{S}}{R} | |
\infty |
Question 6 Explanation:
In maximum power transformation, half of the voltage drops across source resistance, remaining half across the load.
\therefore voltage across source (R)
\begin{aligned} I_{L} R &=\frac{V_{s}}{2} \\ I_{L} &=\frac{V_{s}}{2 R} \end{aligned}
\therefore voltage across source (R)
\begin{aligned} I_{L} R &=\frac{V_{s}}{2} \\ I_{L} &=\frac{V_{s}}{2 R} \end{aligned}
Question 7 |
The switch has been in position 1 for a long time and abruptly changes to position 2 at t=0.
If time t is in seconds, the capacitor voltage V_{C} (in volts) for t \gt 0 is given by
If time t is in seconds, the capacitor voltage V_{C} (in volts) for t \gt 0 is given by
4(1 - exp(-t/0.5)) | |
10 - 6 exp(-t/0.5) | |
4(1 - exp(-t/0.6)) | |
10 - 6 exp(-t/0.6) |
Question 7 Explanation:
at t = 0^{-}, Switch is at position-1
where, \quad V_{c}\left(0^{-}\right)=\frac{10 \times 2}{2+3}=4 \mathrm{V} \ldots(1) \\ \therefore \quad V_{c}(0-)=V_{c}\left(0^{+}\right)=4 \mathrm{V}
\text{at }t = \infty
V_{c}(\infty)=5 \times 2=10 \mathrm{V}
The time constant of the circuit is
\begin{aligned} \tau &=R_{\mathrm{eq}} C_{\mathrm{eq}} \\ &=(4+2) \times 0.1=0.6 \mathrm{sec} \\ \therefore V_{c}(t)=V_{c}(\infty) &+\left[V_{c}\left(0^{+}\right)-V_{c}(\infty)\right] e^{-t / \tau} \\ &=10+(4-10) e^{-t / 0.6} \\ V_{c}(t) &=\left(10-6 e^{-t / 0.6}\right) \mathrm{V} \end{aligned}
where, \quad V_{c}\left(0^{-}\right)=\frac{10 \times 2}{2+3}=4 \mathrm{V} \ldots(1) \\ \therefore \quad V_{c}(0-)=V_{c}\left(0^{+}\right)=4 \mathrm{V}
\text{at }t = \infty
V_{c}(\infty)=5 \times 2=10 \mathrm{V}
The time constant of the circuit is
\begin{aligned} \tau &=R_{\mathrm{eq}} C_{\mathrm{eq}} \\ &=(4+2) \times 0.1=0.6 \mathrm{sec} \\ \therefore V_{c}(t)=V_{c}(\infty) &+\left[V_{c}\left(0^{+}\right)-V_{c}(\infty)\right] e^{-t / \tau} \\ &=10+(4-10) e^{-t / 0.6} \\ V_{c}(t) &=\left(10-6 e^{-t / 0.6}\right) \mathrm{V} \end{aligned}
Question 8 |
The figure shows an RLC circuit with a sinusoidal current source.
At resonance, the ratio |I_{L}|/|I_{R}|, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is ________
At resonance, the ratio |I_{L}|/|I_{R}|, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is ________
0.1 | |
0.3 | |
0.6 | |
0.9 |
Question 8 Explanation:
At resonance (for parallel RLC circuit)
\begin{array}{l} I_{R}=I \\ I_{L}=Q I \angle-90^{\circ} \\ I_{C}=Q I \angle 90^{\circ} \end{array}
For parallel RLC circuit
\begin{aligned} \frac{\left|I_{L}\right|}{\left|I_{R}\right|} &=\frac{I Q}{I}=Q=R \sqrt{\frac{C}{L}} \\ &=10 \sqrt{\frac{10 \times 10^{-6}}{10 \times 10^{-3}}}=0.316 \end{aligned}
\begin{array}{l} I_{R}=I \\ I_{L}=Q I \angle-90^{\circ} \\ I_{C}=Q I \angle 90^{\circ} \end{array}
For parallel RLC circuit
\begin{aligned} \frac{\left|I_{L}\right|}{\left|I_{R}\right|} &=\frac{I Q}{I}=Q=R \sqrt{\frac{C}{L}} \\ &=10 \sqrt{\frac{10 \times 10^{-6}}{10 \times 10^{-3}}}=0.316 \end{aligned}
Question 9 |
The z-parameter matrix for the two-port network shown is
\begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3+2j\omega \end{bmatrix}
where the entries are in \Omega. Suppose Z_{b}(j\omega )=R_{b}+j\omega. Then the value of R_{b} (in \Omega) equals ________
\begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3+2j\omega \end{bmatrix}
where the entries are in \Omega. Suppose Z_{b}(j\omega )=R_{b}+j\omega. Then the value of R_{b} (in \Omega) equals ________
1 | |
2 | |
3 | |
4 |
Question 9 Explanation:
For T-network,
\begin{aligned} Z_{11} &=Z_{a}+Z_{c} \\ Z_{22} &=Z_{b}+Z_{c} \\ \text { and, } \quad Z_{12} &=Z_{21}=Z_{c} \\ \text { Given, } &[Z]=\left[\begin{array}{cc} 2 j \omega & j \omega \\ j \omega & 3+2 j \omega \end{array}\right] \\ \text { Therefore } Z_{12} &=j \omega \\ \text { and, } \quad Z_{22} &=3+2 j \omega \\ &=3+j \omega+j \omega \\ &=Z_{b}+Z_{c} \\ &=R_{b}+j \omega+Z_{c} \\ & R_{b}=3 \Omega \end{aligned}
\begin{aligned} Z_{11} &=Z_{a}+Z_{c} \\ Z_{22} &=Z_{b}+Z_{c} \\ \text { and, } \quad Z_{12} &=Z_{21}=Z_{c} \\ \text { Given, } &[Z]=\left[\begin{array}{cc} 2 j \omega & j \omega \\ j \omega & 3+2 j \omega \end{array}\right] \\ \text { Therefore } Z_{12} &=j \omega \\ \text { and, } \quad Z_{22} &=3+2 j \omega \\ &=3+j \omega+j \omega \\ &=Z_{b}+Z_{c} \\ &=R_{b}+j \omega+Z_{c} \\ & R_{b}=3 \Omega \end{aligned}
Question 10 |
The energy of the signal x(t)=\frac{sin(4\pi t)}{4\pi t} is ________
0 | |
0.25 | |
0.5 | |
0.75 |
Question 10 Explanation:
\begin{aligned} \text { Energy, } E_{x(i)} &=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|X(\omega)|^{2} d \omega \\ &=\frac{1}{2 \pi} \int_{-4 \pi}^{4 \pi}\left(\frac{1}{4}\right)^{2} d \omega \\ &=\frac{1}{2 \pi} \times \frac{1}{16}[8 \pi]=\frac{1}{4}=0.25 \mathrm{J} \end{aligned}
There are 10 questions to complete.