Question 1 |
A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal
voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage
source for all load currents going down to 0 A. When connected to an ideal rheostat,
find the load resistance value at which maximum power is transferred, and the corresponding
load voltage and current.
Short, \infty A, 10 V | |
Open, 4 A, 0 V | |
2.5 \Omega, 4 A, 10 V | |
2.5 \Omega, 4 A, 5 V |
Question 1 Explanation:
Maximum power transistor of VI product is maximum. If draw the the curve, it intersect (10, 4) that will give maximum power. The terminal voltage is 10 V (Load voltage) and current is 4 A (Load current). Load resistance is \frac{10}{4}=2.5\Omega .
Question 2 |
The Thevenin equivalent voltage, V_{TH}, in V (rounded off to 2 decimal places) of the network
shown below, is _______ .
5 | |
7 | |
14 | |
6 |
Question 2 Explanation:
Only voltage source 4V is there and current source 5A is open circuited
From the above circuit,
V_{TH1}=4V
Case-II:
Only current source 5A is there and voltage source 4V is short circuited.
From the above circuit,
V_{TH2}=2 \times 5=10V
By applying superposition theorem,
V_{TH}=V_{TH1}+V_{TH2}=10+4=14V
From the above circuit,
V_{TH1}=4V
Case-II:
Only current source 5A is there and voltage source 4V is short circuited.
From the above circuit,
V_{TH2}=2 \times 5=10V
By applying superposition theorem,
V_{TH}=V_{TH1}+V_{TH2}=10+4=14V
Question 3 |
x_R\; and \; x_A are, respectively, the rms and average values of x(t) = x(t - T), and similarly,
y_R\; and \; y_A are, respectively, the rms and average values of y(t) = kx(t). k, \;T are
independent of t. Which of the following is true?
y_A=kx_A;\; y_R=kx_R | |
y_A=kx_A;\; y_R\neq kx_R | |
y_A \neq kx_A;\; y_R= kx_R | |
y_A \neq kx_A;\; y_R\neq kx_R |
Question 3 Explanation:
Given that,
\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}
\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}
Question 4 |
The current flowing in the circuit shown below in amperes is _____
0 | |
1 | |
2 | |
4 |
Question 4 Explanation:
By Millman'e theorem,
E=\frac{\frac{200}{50}+\frac{160}{40}-\frac{100}{25}-\frac{80}{20}}{\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}}=0V
\frac{1}{R}=\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}
Simplified circuit,
\therefore \;\;I=0A
Question 5 |
In the circuit shown below, the value of capacitor C required for maximum power to be
transferred to the load is
1nF | |
1\muF | |
1mF | |
10mF |
Question 5 Explanation:
The frequency at which the load is resistive and it is equal to 0.5\Omega i.e. The load is resistive means, the imaginary part of load is equal to zero and real part is equal to 0.5\Omega.
Z_{load}=\frac{1 \times \frac{1}{Cs}}{1 + \frac{1}{Cs}}+Ls
\;\;=\frac{1}{1+Cs}+Ls
\;\; =\frac{(1-Cs)}{1-C^2s^2}+Ls
Put, s=j\omega;
Z_{load}=\frac{1-j\omega C}{1+C^2\omega ^2}+j\omega L
\;\;\;=\frac{1}{1+C^2\omega ^2}+j\omega \left ( L-\frac{C}{(1+C^2\omega ^2)} \right )
Real part of the
Z_{load}=\frac{1}{1+C^2\omega ^2}=0.5
Putting, \omega =100 r/s
we get, C=10mF
Question 6 |
For the network given in figure below, the Thevenin's voltage V_{ab} is
-1.5 V | |
-0.5 V | |
0.5V | |
1.5V |
Question 6 Explanation:
Consider the following circuit,
After rearrangement, we get,
From circuit using KCL,
\frac{V_{ab}+30}{15}+\frac{V_{ab}-8}{5}=0
V_{ab}+30+3V_{ab}-24=0
V_{ab}=-1.5V
After rearrangement, we get,
From circuit using KCL,
\frac{V_{ab}+30}{15}+\frac{V_{ab}-8}{5}=0
V_{ab}+30+3V_{ab}-24=0
V_{ab}=-1.5V
Question 7 |
For the given 2-port network, the value of transfer impedance Z_{21} in ohms is_______
1 | |
2 | |
3 | |
4 |
Question 7 Explanation:
Where, R_A=1\Omega
R_B=1\Omega
R_C=\frac{1}{2}\Omega
After rearrangement consider the following circuit,
From the circuit diagram, we get,
Z_{21}=\frac{V_2}{I_1}=3\Omega
Question 8 |
In the circuit shown below, the maximum power transferred to the resistor R is _______ W.
2.5 | |
1.5 | |
6 | |
3 |
Question 8 Explanation:
To get R_{th} and V_{th}, consider the following steps.
Case-1: For R_{th}
For V_{th}
Applying KCL at node,
\frac{V_{th}-5}{5}+\frac{V_{th}+16}{5}=0
2V_{th}=-11
V_{th}=-5.5 V
Maximum power transferred,
P_{max}=\frac{V_{th}^2}{4R_{th}}= \frac{5.5^2}{4 \times 2.5}=3.025W
Case-1: For R_{th}
For V_{th}
Applying KCL at node,
\frac{V_{th}-5}{5}+\frac{V_{th}+16}{5}=0
2V_{th}=-11
V_{th}=-5.5 V
Maximum power transferred,
P_{max}=\frac{V_{th}^2}{4R_{th}}= \frac{5.5^2}{4 \times 2.5}=3.025W
Question 9 |
In a linear two-port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short-circuited. When 5 V is applied to Port 1, a current of 1.25 A flows through a 1 \Omega resistance connected across Port 2. When 3 V is applied to Port 1, the current (in Ampere) through a 2 \Omega resistance connected across Port 2 is _______.
0.225 | |
0.545 | |
0.845 | |
1.475 |
Question 9 Explanation:
V_1=10V, I_2=4 A, V_2=0\;\;Cond...(i)
V_1=5V, I_2=1.25 A, V_2=1.25 \times 1=1.25V\;\;Cond...(ii)
V_1=3V, I_2=? , R=2\Omega \;\;Cond...(iii)
as we know from ABCD parameter,
V_1=AV_2-BI_2
I_1=CV_2-DI_2
From condition (i),
10=A(0)-B(4)
B=-\frac{10}{4}
From condition (ii),
5=A(1.25)-\left ( -\frac{10}{4} \right ) \times 1.25
A=\frac{\left ( 5-\frac{12.5}{4} \right )}{1.25}=1.5
From condition (iii),
3=1.5(2I)-\left ( -\frac{10}{4} \right )\times I
3=3I+2.5I
I=0.545 A
Question 10 |
The circuit s hown in the figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by v(t)=12\sin t. If the circuit is in steady state, then the rms value of the current (in Ampere) flowing in the circuit is ______ .
7 | |
10 | |
15 | |
14 |
Question 10 Explanation:
i(t)=\frac{8}{1}+\frac{12 \sin t}{\sqrt{1+1}}=8+6\sqrt{2} \sin t
i_{rms}(t)=\sqrt{8^2+\frac{1}{2}(6\sqrt{2} )^2}=10A
i_{rms}(t)=\sqrt{8^2+\frac{1}{2}(6\sqrt{2} )^2}=10A
There are 10 questions to complete.
