Solved GATE Paper

GATE 2009 Electric Circuits Questions and Solutions

Instructor: Prof. Mithun Mondal Institution: BITS Pilani Subject: Electric Circuits
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Question 01

Question 1

In the circuit shown in the figure , what value of \(R_L\) maximizes the power delivered to \(R_L\)?

GATE 2009 Electric Circuits Q1 circuit diagram
Circuit for GATE 2009 Electric Circuits Q1
  1. \(2.4\,\Omega\)
  2. \(8/3\,\Omega\)
  3. \(4\,\Omega\)
  4. \(6\,\Omega\)

Solution

  • Condition: For maximum power transfer (MPT), the load resistance must equal the Thevenin resistance, \(R_L = R_{th}\).
  • Finding \(R_{th}\): To find \(R_{th}\), the independent source (100V) must be deactivated (short-circuited). A test source method or finding \(V_{oc}/I_{sc}\) is typically used for circuits with dependent sources.
  • Result: The calculated Thevenin resistance for this specific circuit is \(R_{th} = 4\,\Omega\).
  • Value: Therefore, \(R_L = 4\,\Omega\).
C
Final Answer
Correct answer: (3) \(4\,\Omega\).
Question 02

Question 2

The current \(I\) in the circuit shown in the figure is:

GATE 2009 Electric Circuits Q2 circuit diagram
Circuit for GATE 2009 Electric Circuits Q2
  1. \(-j1\,A\)
  2. \(j1\,A\)
  3. \(0\,A\)
  4. \(20\,A\)

Solution

  • Impedance Calculation: Given \(\omega=10^3\,rad/s\).
    • Inductive Reactance: \(X_L = \omega L = 10^3 \times 20 \times 10^{-3} = 20\,\Omega\). Impedance \(Z_L = j20\,\Omega\).
    • Capacitive Reactance: \(X_C = \frac{1}{\omega C} = \frac{1}{10^3 \times 50 \times 10^{-6}} = 20\,\Omega\). Impedance \(Z_C = -j20\,\Omega\).
  • Nodal Analysis: Apply Nodal analysis at the top node (\(V_A\)) with the source \(V_S = 20\angle 0^\circ\,V\).
    Equation
    \[\frac{V_A - 20}{Z_L} + \frac{V_A}{R} + \frac{V_A}{Z_C} = 0\]
    Equation
    \[\frac{V_A - 20}{j20} + \frac{V_A}{1} + \frac{V_A}{-j20} = 0\]
  • Solving for \(V_A\): The terms \(\frac{V_A}{j20}\) and \(\frac{V_A}{-j20}\) cancel out.
    Equation
    \[V_A - \frac{20}{j20} = 0 \Rightarrow V_A - (-j1) = 0 \Rightarrow V_A = -j1\,V\]
  • Finding Current \(I\): The current \(I\) is the current flowing down through the \(1\,\Omega\) resistor, \(R\):
    Equation
    \[I = \frac{V_A}{R} = \frac{-j1\,V}{1\,\Omega} = -j1\,A\]
A
Final Answer
Correct answer: (1) \(-j1\,A\).
Question 03

Question 3

The condition for maximum power transfer to a load impedance \(Z_L = R_L + jX_L\) from a source with internal impedance \(Z_S = R_S + jX_S\) is:

  1. \(R_L = R_S\) and \(X_L = X_S\)
  2. \(R_L = R_S\) and \(X_L = -X_S\)
  3. \(R_L = |Z_S|\) and \(X_L = 0\)
  4. \(Z_L = R_S\)

Solution

  • For a complex source impedance, maximum power is transferred when the load impedance is the conjugate of the source impedance, i.e., \(Z_L = Z_S^*\).
B
Final Answer
Correct answer: (2) \(R_L = R_S\) and \(X_L = -X_S\).
Question 04

Question 4

A two-port network is reciprocal if its admittance parameters satisfy which of the following conditions?

  1. \(y_{11} = y_{22}\)
  2. \(y_{12} = -y_{21}\)
  3. \(y_{12} = y_{21}\)
  4. \(y_{11}y_{22} - y_{12}y_{21} = 1\)

Solution

  • A two-port network is considered reciprocal if the ratio of the output current (Port 2) to the input voltage (Port 1) equals the ratio of the input current (Port 1) to the output voltage (Port 2) under appropriate open/short circuit conditions. In terms of the Y-parameters (Admittance parameters), this condition is \(y_{12} = y_{21}\).
  • The condition \(y_{11} = y_{22}\) indicates a symmetric network.
C
Final Answer
Correct answer: (3) \(y_{12} = y_{21}\).
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GATE Electric Circuits