Solved GATE Paper

GATE 2012 Electric Circuits Questions and Solutions

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

Question 1

Assuming both the voltage sources are in phase, the value of \(R\) for which maximum power is transferred from circuit A to circuit B in the figure is:

GATE 2012 Electric Circuits Q1 circuit diagram
Circuit for GATE 2012 Electric Circuits Q1
  1. \(0.8\,\Omega\)
  2. \(1.4\,\Omega\)
  3. \(2\,\Omega\)
  4. \(2.8\,\Omega\)

Solution

  • Loop Current: Both sources are in phase, so the net driving voltage is \(10\,\text{V} - 3\,\text{V} = 7\,\text{V}\) acting on the series resistance \(2\,\Omega + R\):
    Equation
    \[I = \frac{7}{R+2}\]
  • Power Delivered to Circuit B: Circuit B contains the \(3\,\text{V}\) source as well as \(R\), so what crosses the interface is
    Equation
    \[P = 3I + I^2 R = \frac{21}{R+2} + \frac{49R}{(R+2)^2}\]
    Maximising \(I^2R\) on its own would give \(R = 2\,\Omega\); that is the power burnt in \(R\), not the power transferred from A to B.
  • Maximization: Differentiating and setting \(\frac{dP}{dR} = 0\):
    Equation
    \[-\frac{21}{(R+2)^2} + \frac{49(2-R)}{(R+2)^3} = 0 \Rightarrow -21(R+2) + 49(2-R) = 0\]
    Equation
    \[-70R + 56 = 0 \Rightarrow R = 0.8\,\Omega\]
A
Final Answer
Correct answer: (1) \(0.8\,\Omega\).
Question 02

Question 2

The impedance looking into nodes 1 and 2 in the given circuit in the figure is:

GATE 2012 Electric Circuits Q2 circuit diagram
Circuit for GATE 2012 Electric Circuits Q2
  1. \(50\,\Omega\)
  2. \(100\,\Omega\)
  3. \(5\,k\Omega\)
  4. \(10.1\,k\Omega\)

Solution

  • Method: Use the test source method. Apply a test voltage \(V_{\text{test}} = 1\,\text{V}\) across terminals 1 and 2 and find the current \(I_{\text{test}}\). \(Z_{\text{th}} = V_{\text{test}} / I_{\text{test}}\).
  • Calculate \(i_b\): The current \(i_b\) flows through the series \(9\,k\Omega\) and \(1\,k\Omega\) resistors.
    Equation
    \[i_b = \frac{V_{\text{test}}}{10\,k\Omega} = \frac{1\,\text{V}}{10 \times 10^3\,\Omega} = 10^{-4}\,\text{A}\]
  • Calculate \(I_{\text{test}}\): \(I_{\text{test}}\) is the sum of currents leaving the \(1\,\text{V}\) source, which flows through the \(100\,\Omega\) resistor and the dependent current source path (as per the implied simplification in the source solution).
    Equation
    \[I_{\text{test}} = 200\,i_b = 200 \times 10^{-4}\,\text{A} = 0.02\,\text{A}\]
  • Impedance:
    Equation
    \[Z_{\text{th}} = \frac{V_{\text{test}}}{I_{\text{test}}} = \frac{1\,\text{V}}{0.02\,\text{A}} = 50\,\Omega\]
A
Final Answer
Correct answer: (1) \(50\,\Omega\).
Question 03

Question 3

In the circuit shown in the figure , the three voltmeter readings are \(V_{1}=220\,\text{V}\), \(V_{2}=122\,\text{V}\), and \(V_{3}=136\,\text{V}\). The power factor of the load is:

GATE 2012 Electric Circuits Q3 circuit diagram
Circuit for GATE 2012 Electric Circuits Q3
  1. \(0.45\)
  2. \(0.5\)
  3. \(0.55\)
  4. \(0.6\)

Solution

  • Law of Cosines: The voltages form a triangle where \(V_1\) is the total voltage, and \(V_2\) (voltage across \(R\)) and \(V_3\) (voltage across the load) are the legs. The phase angle (\(\phi\)) between \(V_2\) and \(V_3\) gives the power factor angle of the load.
    Equation
    \[V_1^2 = V_2^2 + V_3^2 + 2V_2 V_3 \cos(\phi)\]
    Equation
    \[(220)^2 = (122)^2 + (136)^2 + 2(122)(136) \cos(\phi)\]
    Equation
    \[48400 = 14884 + 18496 + 33184 \cos(\phi)\]
  • Power Factor: Solving for the power factor \(\cos(\phi)\):
    Equation
    \[\cos(\phi) = \frac{48400 - 33380}{33184} \approx 0.45\]
A
Final Answer
Correct answer: (1) \(0.45\).
Question 04

Question 4

(Linked to Question 3) In the circuit shown, \(V_{1}=220\,\text{V}\), \(V_{2}=122\,\text{V}\), and \(V_{3}=136\,\text{V}\). If \(R_{L}=5\,\Omega\), the approximate power consumption in the load is:

  1. \(700\,W\)
  2. \(750\,W\)
  3. \(800\,W\)
  4. \(850\,W\)

Solution

  • Load Impedance Magnitude: Using \(\cos(\phi) \approx 0.45\) from Question 3, the load impedance magnitude \(|Z_L|\) is:
    Equation
    \[|Z_L| = \frac{R_L}{\cos(\phi)} = \frac{5\,\Omega}{0.45} \approx 11.11\,\Omega\]
  • Load Current: The RMS current through the load is \(I = V_3 / |Z_L|\).
    Equation
    \[I = \frac{136\,\text{V}}{11.11\,\Omega} \approx 12.24\,\text{A}\]
  • Load Power: Power consumed by the load (\(P_L\)) is the power dissipated in its resistive part \(R_L\):
    Equation
    \[P_L = I^2 R_L \approx (12.24\,\text{A})^2 \times 5\,\Omega \approx 750\,\text{W}\]
B
Final Answer
Correct answer: (2) \(750\,W\).
Question 05

Question 5

The average power delivered to an impedance \((4-j3)\,\Omega\) by a current \(5 \cos(100\pi t + 100^\circ)\,A\) is:

  1. \(44.2\,W\)
  2. \(50\,W\)
  3. \(62.5\,W\)
  4. \(125\,W\)

Solution

  • Formula: Average power delivered to an impedance \(Z = R + jX\) is \(P = I_{\text{rms}}^2 R\).
  • Parameters: The resistive part of the impedance is \(R = 4\,\Omega\). The peak current is \(I_m = 5\,\text{A}\).
  • RMS Current:
    Equation
    \[I_{\text{rms}} = \frac{I_m}{\sqrt{2}} = \frac{5}{\sqrt{2}}\,\text{A}\]
  • Calculation:
    Equation
    \[P = I_{\text{rms}}^2 R = \left(\frac{5}{\sqrt{2}}\right)^2 \times 4 = \frac{25}{2} \times 4 = 50\,\text{W}\]
B
Final Answer
Correct answer: (2) \(50\,W\).
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GATE Electric Circuits