Question 1 |
In the circuit shown below, the Thevenin voltage V_{TH} is
2.4 V | |
2.8 V | |
3.6 V | |
4.5 V |
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?
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?
0.5 A | |
1.0 A | |
2.0 A | |
2.5 A |
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}
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 _________.
The Thevenin equivalent resistance (in \Omega) across P - Q is _________.
0 | |
-1 | |
-2 | |
1 |
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}
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
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 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}
\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 __________
0.2 | |
0.6 | |
0.8 | |
1 |
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.