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.