GATE 2011 Electric Circuits Questions and Solutions

Question 01

Question 1

An RLC circuit with relevant data is given in the figure . The power dissipated in the resistor \(R\) is: (\(V_S = 1\angle 0^\circ V\) and \(I_{RL} = \sqrt{2}\angle -\pi/4 A\) are given.)

GATE 2011 Electric Circuits Q1 circuit diagram — Circuit for GATE 2011 Electric Circuits Q1

\(0.5\,W\)
\(1\,W\)
\(\sqrt{2}\,W\)
\(2\,W\)

Solution

Method: \(R\) and \(L\) carry the same current \(I_{RL}\) and the inductor is lossless, so the whole of the real power entering that branch ends up in \(R\). The branch sits directly across \(V_S\), hence \(P_R = \operatorname{Re}\{V_S I_{RL}^{*}\}\). The capacitor branch draws only reactive power and does not enter the calculation.
Power Calculation: With \(V_S = 1\angle 0^\circ\,V\) and \(I_{RL} = \sqrt{2}\angle -\pi/4\,A\), the angle between voltage and branch current is \(\pi/4\):
Equation
\[P_R = |V_S| |I_{RL}| \cos(\pi/4) = 1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} = 1\,W\]
Check: The branch impedance is \(Z_{RL} = V_S / I_{RL} = \frac{1\angle 0^\circ}{\sqrt{2}\angle -\pi/4} = \frac{1}{\sqrt{2}}\angle 45^\circ\,\Omega\), so \(R = \frac{1}{\sqrt{2}}\cos 45^\circ = 0.5\,\Omega\) and \(P_R = |I_{RL}|^2 R = 2 \times 0.5 = 1\,W\).

Final Answer

Correct answer: (2) \(1\,W\).

Question 02

Question 2

The r.m.s value of the current \(i(t)\) in the circuit shown in the figure is:

GATE 2011 Electric Circuits Q2 circuit diagram — Circuit for GATE 2011 Electric Circuits Q2

\(\frac{1}{2}\,A\)
\(\frac{1}{\sqrt{2}}\,A\)
\(1\,A\)
\(2\,A\)

Solution

Source and Frequency: The source voltage is \(V_S(t) = 1.0 \sin(t)\,V\), which gives a peak voltage \(V_m = 1\,V\) and angular frequency \(\omega = 1\,rad/s\).
Impedance Calculation: The inductor and the capacitor sit in series with the resistor, so their reactances add:
Equation
\[Z_{LC} = j\omega L + \frac{1}{j\omega C} = j(1)(1) + \frac{1}{j(1)(1)} = j1 - j1 = 0\]
Total Impedance: The L–C section is at series resonance at \(\omega = 1\,rad/s\), so its net reactance is zero and it behaves as a short. The total impedance seen by the voltage source is the \(1\,\Omega\) resistor alone.
Equation
\[Z_{total} = R_{series} + Z_{LC} = 1\,\Omega + 0\,\Omega = 1\,\Omega\]
Instantaneous Current: The current is \(i(t) = \frac{V_S(t)}{Z_{total}} = \frac{1.0 \sin(t)\,V}{1\,\Omega} = \sin(t)\,A\).
RMS Value: For a sinusoidal current \(i(t) = I_m \sin(\omega t)\), the RMS value is \(I_{rms} = \frac{I_m}{\sqrt{2}}\).
Equation
\[I_{rms} = \frac{1\,A}{\sqrt{2}} = \frac{1}{\sqrt{2}}\,A\]

Final Answer

Correct answer: (2) \(\frac{1}{\sqrt{2}}\,A\).

Question 03

Question 3

In the circuit shown in the figure , the current \(I\) is equal to:

GATE 2011 Electric Circuits Q3 circuit diagram — Circuit for GATE 2011 Electric Circuits Q3

\(14\angle 0^\circ\,A\)
\(2.0\angle 0^\circ\,A\)
\(2.8\angle 0^\circ\,A\)
\(3.2\angle 0^\circ\,A\)

Solution

Method: The three \(6\,\Omega\) resistors form a Delta (\(\Delta\)) network. Convert this to an equivalent Wye (Y) network using \(\Delta\)-Y transformation.
Y-Resistor Value: Since all resistors in the \(\Delta\) are equal (\(R_\Delta = 6\,\Omega\)), the equivalent Y-resistor value is \(R_Y = \frac{R_\Delta \times R_\Delta}{3 R_\Delta} = \frac{6 \times 6}{3 \times 6} = 2\,\Omega\).
Simplified Circuit: The circuit simplifies to a series impedance \(Z_{series} = 2\,\Omega\) followed by a parallel combination: \(Z_{parallel} = (R_Y + j4\,\Omega) || (R_Y - j4\,\Omega)\).
Equivalent Impedance:
Equation
\[Z_{parallel} = \frac{(2+j4)(2-j4)}{(2+j4)+(2-j4)} = \frac{2^2 - (j4)^2}{4} = \frac{4 + 16}{4} = 5\,\Omega\]
Total Impedance: The total impedance seen by the source is:
Equation
\[Z_{eq} = 2\,\Omega + Z_{parallel} = 2\,\Omega + 5\,\Omega = 7\,\Omega\]
Current \(I\):
Equation
\[I = \frac{V_S}{Z_{eq}} = \frac{14\angle 0^\circ\,V}{7\,\Omega} = 2\angle 0^\circ\,A\]

Final Answer

Correct answer: (2) \(2.0\angle 0^\circ\,A\).