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}
Question 6
For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a-b is ________.
A
12
B
10
C
8
D
16
GATE EC 2015-SET-3   Network Theory
Question 6 Explanation: 
\begin{aligned} V_{Th}&=V_{6\;\omega}\\ \frac{V_{A}-12}{3}+\frac{V_{A}}{6} &=V_{6 \Omega} \\ V_{A}\left[\frac{1}{3}+\frac{1}{6}\right] &=1+4 \\ V_{A} \frac{3}{6} &=5 \\ V_{T h} &=V_{6 \Omega}=V_{A}=10 \mathrm{V} \end{aligned}

Question 7
In the circuit shown, the Norton equivalent resistance (in \Omega) across terminals a-b is _______.

A
1.3
B
0.8
C
1.9
D
2.2
GATE EC 2015-SET-2   Network Theory
Question 7 Explanation: 


\begin{aligned} I&=\frac{V_{0}}{4} \\ I_{1}&=\frac{V_{0}}{2}\\ \text{Applying KCL}\\ \frac{V_{0}-4 I}{2}+\frac{V_{0}}{2}+\frac{V_{0}}{4}&=I_{0}\\ \text{From here}\\ V_{0} \cdot \frac{3}{4} &=I_{0} \\ R_{N} &=\frac{V_{0}}{I_{0}}=\frac{4}{3}=1.33 \Omega \end{aligned}
Question 8
In the given circuit, the maximum power (in Watts) that can be transferred to the load R_{L} is_______.

A
1
B
1.6
C
2
D
2.5
GATE EC 2015-SET-1   Network Theory
Question 8 Explanation: 


For maximum power transfer
\begin{aligned} R_{L} &=\left|Z_{T h}\right|=|2 \| j 2| \\ &=\left|\frac{2 \times j 2}{2+j 2}\right|=1.414 \Omega \\ V_{T h} &=\frac{8 \angle 90^{\circ}}{2+j 2}=2.828 \angle 45^{\circ} \end{aligned}


\begin{aligned} I &=\frac{2.828 \angle 45^{\circ}}{1.414 \angle 45^{\circ}+1.414}=1.08 \angle 22.5^{\circ} \\ \text { Power } &=I^{2} R=(1.08)^{2} \times \sqrt{2}=1.649 \mathrm{W} \end{aligned}
Question 9
In the circuit shown in the figure, the angular frequency \omega (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b' is purely resistive, is _____.
A
1
B
2
C
3
D
4
GATE EC 2014-SET-3   Network Theory
Question 9 Explanation: 


Finding Z_{N}
\begin{aligned} Z_{b b}^{\prime} &=\frac{1 \times j 0.5 \omega}{1+j 0.5 \omega}+\frac{1}{j \omega}=\frac{j \omega}{2+j \omega}+\frac{1}{j \omega} \\ \text { or } Z_{b b}^{\prime} &=\frac{2-\omega^{2}+j \omega}{2 j \omega-\omega^{2}}&\ldots(i) \end{aligned}
Rationalizing equation (i), we get,
\begin{aligned} Z_{b b}^{\prime} &=\frac{\left(2-\omega^{2}\right)+j \omega}{2 j \omega-\omega^{2}} \times \frac{-\omega^{2}-j 2 \omega}{-\omega^{2}-j 2 \omega} \\ &=-\frac{2 \omega^{2}+\omega^{4}+2 \omega^{2}}{\omega^{4}+2 \omega^{2}}+j \frac{\left(\omega^{3}-4 \omega\right)}{\omega^{4}+2 \omega^{2}} \end{aligned}
In order to have a purely resistive impedance Z_{\text {bb }}^{\prime} , the imaginary part of equation (ii) will be equaled to zero.
\begin{aligned} &\therefore \frac{-4 \omega+\omega^{3}}{\omega^{4}+2 \omega^{2}}=0\\ &\text{or}\quad \omega^{3}=4 \omega \\ &\text{or}\quad \omega=\sqrt{4}=2 \mathrm{rad} / \mathrm{sec} \end{aligned}
Question 10
In the figure shown, the value of the current I (in Amperes) is_____.
A
0
B
0.25
C
0.5
D
1
GATE EC 2014-SET-3   Network Theory
Question 10 Explanation: 


Using super position theorem, when 5 V source acting alone, we get


I_{1}=\frac{V}{R_{\mathrm{eq}}}=\frac{5}{10+5+5}=\frac{1}{4} \mathrm{A}\quad \ldots(i)
When 1 A source acting alone, we get


I_{2}=\frac{1 \times 5}{5+10+5}=\frac{5}{20}=\frac{1}{4} \mathrm{A}\quad \ldots(ii)
Therefore,
I=I_{1}+I_{2}=\frac{1}{2} A=0.5 \mathrm{A}
There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.