# Network Theorems

 Question 1
In the circuit shown below, the Thevenin voltage $V_{TH}$ is A 2.4 V B 2.8 V C 3.6 V D 4.5 V
GATE EC 2020   Network Theory
Question 1 Explanation:
By applying the Source Transformation Question 2
Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure).
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1? A 0.5 A B 1.0 A C 2.0 A D 2.5 A
GATE EC 2019   Network Theory
Question 2 Explanation:
According to reciprocity theorem,
In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged.
$\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}$

 Question 3
Consider the circuit shown in the figure. The Thevenin equivalent resistance (in $\Omega$) across P - Q is _________.
 A 0 B -1 C -2 D 1
GATE EC 2017-SET-2   Network Theory
Question 3 Explanation:
The equivalent circuit to calculate the Thevenin equivalent resistance ($R_{Th}$) is as follows: lt can be further reduced as follows:  By applying KVL in the Loop L
\begin{aligned} V_{x} &=3 i_{0}+\left(1-i_{0}\right)=2 i_{0}+1 \\ \text { Also, } \quad V_{x} &=i_{0}(1 \Omega) \end{aligned}
\begin{aligned} \text { So, } 2 i_{0}+1&=i_{0} \\ i_{0} &=-1 \mathrm{A} \\ V_{x} &=-1 \mathrm{V}\\ \text { So, } \quad R_{\mathrm{Th}}&=\frac{V_{x}}{1 \mathrm{A}}=-1 \Omega \end{aligned}
 Question 4
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
 A $\frac{V_{S}}{4R}$ B $\frac{V_{S}}{2R}$ C $\frac{V_{S}}{R}$ D $\infty$
GATE EC 2016-SET-2   Network Theory
Question 4 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}
 Question 5
In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is __________ A 0.2 B 0.6 C 0.8 D 1
GATE EC 2016-SET-1   Network Theory
Question 5 Explanation: For maximum power transfer,
$\begin{array}{l} R=R_{T h} \\ V_{0}=5 \times \frac{2 \mathrm{k} \Omega}{5 \mathrm{k} \Omega}=2 \mathrm{V} \end{array}$
From output loop, $V_{T H}=100 \times 2 \times \frac{40 \mathrm{k} \Omega}{50 \mathrm{k} \Omega}$
$\begin{array}{l} V_{\mathrm{TH}}=160 \mathrm{V}\\ \text { and } R_{\mathrm{TH}}=10 \mathrm{k} \Omega|| 40 \mathrm{k} \Omega=\frac{10 \times 40}{50}=8 \mathrm{k} \Omega \\ \therefore \text { Maximum power }=\frac{V^{2}_{\mathrm{Th}}}{4 R_{\mathrm{Th}}}=\frac{160 \times 160}{4 \times 8000}=0.8 \mathrm{W} \end{array}$

There are 5 questions to complete.