# Network Theorems

 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.
 A Short, $\infty$ A, 10 V B Open, 4 A, 0 V C 2.5 $\Omega$, 4 A, 10 V D 2.5 $\Omega$, 4 A, 5 V
GATE EE 2020   Electric Circuits
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 _______ . A 5 B 7 C 14 D 6
GATE EE 2020   Electric Circuits
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,
$V_{TH}=V_{TH1}+V_{TH2}=10+4=14V$
 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?
 A $y_A=kx_A;\; y_R=kx_R$ B $y_A=kx_A;\; y_R\neq kx_R$ C $y_A \neq kx_A;\; y_R= kx_R$ D $y_A \neq kx_A;\; y_R\neq kx_R$
GATE EE 2020   Electric Circuits
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}
 Question 4
The current flowing in the circuit shown below in amperes is _____ A 0 B 1 C 2 D 4
GATE EE 2019   Electric Circuits
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 A 1nF B 1$\mu$F C 1mF D 10mF
GATE EE 2017-SET-2   Electric Circuits
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 A -1.5 V B -0.5 V C 0.5V D 1.5V
GATE EE 2017-SET-2   Electric Circuits
Question 6 Explanation:
Consider the following circuit, After rearrangement, we get, From circuit using KCL,
$\frac{V_{ab}+30}{15}+\frac{V_{ab}-8}{5}=0$
$V_{ab}+30+3V_{ab}-24=0$
$V_{ab}=-1.5V$
 Question 7
For the given 2-port network, the value of transfer impedance $Z_{21}$ in ohms is_______ A 1 B 2 C 3 D 4
GATE EE 2017-SET-2   Electric Circuits
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. A 2.5 B 1.5 C 6 D 3
GATE EE 2017-SET-1   Electric Circuits
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,
$P_{max}=\frac{V_{th}^2}{4R_{th}}= \frac{5.5^2}{4 \times 2.5}=3.025W$
 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 _______.
 A 0.225 B 0.545 C 0.845 D 1.475
GATE EE 2015-SET-1   Electric Circuits
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 ______ . A 7 B 10 C 15 D 14
GATE EE 2015-SET-1   Electric Circuits
Question 10 Explanation:
$i(t)=\frac{8}{1}+\frac{12 \sin t}{\sqrt{1+1}}=8+6\sqrt{2} \sin t$
$i_{rms}(t)=\sqrt{8^2+\frac{1}{2}(6\sqrt{2} )^2}=10A$
There are 10 questions to complete. 