GATE 2011 Electric Circuits Questions and Solutions

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.)

1. \(0.5\,W\)
2. \(1\,W\)
3. \(\sqrt{2}\,W\)
4. \(2\,W\)

Solution

• Method: The power dissipated in the resistor \(R\) is \(P_R = |I_{RL, rms}|^2 R\). Alternatively, the solution uses the total power \(P_S\) assuming the inductor and capacitor are lossless.
• Power Calculation (from source's method): The power supplied by the source \(P_S\) is calculated using the total source current magnitude \(|I_S|\) and the phase difference \(\phi\):
  Equation

  \[P_S = |V_S| |I_S| \cos(\phi) = (1) \times (\sqrt{2}) \times \cos(\pi/4) = 1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} = 1\,W\]
  (Assuming \(|I_S| = |I_{RL}| = \sqrt{2}\,A\) and the phase angle of the total current is \(\pi/4\) for the power calculation.)

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

1. \(\frac{1}{2}\,A\)
2. \(\frac{1}{\sqrt{2}}\,A\)
3. \(1\,A\)
4. \(2\,A\)

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 parallel LC branch impedance (\(Z_{LC}\)) is:
  Equation

  \[Z_{LC} = j\omega L + \frac{1}{j\omega C} = j(1)(1) + \frac{1}{j(1)(1)} = j1 - j1 = 0\]
• Total Impedance: Since the parallel LC branch is a short circuit (\(Z_{LC}=0\)), the total impedance seen by the voltage source is only the series resistor (\(1\,\Omega\)).
  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

  \[I_{rms} = \frac{1\,A}{\sqrt{2}} = \frac{1}{\sqrt{2}}\,A\]

Note: the L–C section is in series and reaches resonance at \(\omega=1\,\mathrm{rad/s}\), so its net reactance is zero and it behaves as a short; the total impedance is the \(1\,\Omega\) resistor alone.

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

1. \(14\angle 0^\circ\,A\)
2. \(2.0\angle 0^\circ\,A\)
3. \(2.8\angle 0^\circ\,A\)
4. \(3.2\angle 0^\circ\,A\)

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

  \[I = \frac{V_S}{Z_{eq}} = \frac{14\angle 0^\circ\,V}{7\,\Omega} = 2\angle 0^\circ\,A\]