GATE 2016 Electric Circuits Questions and Solutions

Consider the network shown below with \(R_1=1\Omega\), \(R_2=2\Omega\) and \(R_3=3\Omega\). The network is connected to a constant voltage source of 11 V. The magnitude of the current (in amperes) delivered by the voltage source is _____.

Solution

Find equivalent resistance \(R_{eq}\). (Following the solution's formula):

In the circuit shown in the figure, the magnitude of the current (in amperes) through \(R_2\) is _____.

Solution

Nodal analysis at node \(v_A\).

The voltage \(v_x\) is across \(R_3\). Using voltage divider:

Substitute \(v_x\) into the nodal equation:

Current through \(R_2 = \frac{v_A}{R_2 + R_3} = \frac{40}{3 + 5} = 5~A\).

In the figure shown, the current i (in ampere) is _____.

Solution

Nodal analysis at center node \(V_1\):

Using KCL from solution: \(i = V_1 - 5 = 4 - 5 = -1A\).

In the circuit shown, the Norton equivalent resistance (in \(\Omega\)) across terminals a-b is _____.

Solution

Apply \(V_{dc}\) at a-b, find \(I_{dc}\). KCL at node a (Vdc), using \(4I = V_{dc}\) (since \(I = V_{dc}/4\)):

In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is _____.

Solution

\(P_{max} = V_{TH}^2 / (4R_{TH})\). Find \(v_g\) from the left circuit using voltage divider:

Find \(V_{TH}\) for the right circuit (open-circuit voltage across R):

Find \(R_{TH}\) (looking into R terminals, \(v_g\) depends on source, but source is independent and deactivated, so \(v_g=0\), dependent source is 0V \(\to\) short):

Consider a two-port network with the transmission matrix: \(T = \begin{bmatrix} A & B \\ C & D \end{bmatrix}\). If the network is reciprocal, then

1. \(T^{-1} = T\)
2. \(T^2 = T\)
3. Determinant(T) = 0
4. Determinant(T) = 1

Solution

For reciprocity, \(AD - BC = 1\), which is the determinant of the T matrix.

An independent voltage source in series with an impedance \(Z_S = R_S + jX_S\) delivers a maximum average power to a load impedance \(Z_L\) when

1. \(Z_L = R_S + jX_S\)
2. \(Z_L = R_S\)
3. \(Z_L = jX_S\)
4. \(Z_L = R_S - jX_S\)

Solution

The Maximum Power Transfer Theorem states that for maximum average power transfer... the load impedance must be the complex conjugate of the source impedance: \(Z_L = Z_S^*\). If \(Z_S = R_S + jX_S\) then \(Z_L = R_S - jX_S\).

In the AC network shown in the figure, the phasor voltage \(V_{AB}\) (in Volts) is (No image provided in PDF for this question)

1. 0
2. \(5\angle 30^\circ\)
3. \(12.5\angle 30^\circ\)
4. \(17\angle 30^\circ\)

Solution

(Based on the solution) The voltage \(V_{AB}\) is across a parallel combination, driven by a current source \(I = 5\angle 30^\circ~A\). Impedance of left branch \(Z_1 = 5 - j3~\Omega\). Impedance of right branch \(Z_2 = 5 + j3~\Omega\). Equivalent impedance \(Z_{eq} = Z_1 || Z_2 = \frac{(5 - j3)(5 + j3)}{(5 - j3) + (5 + j3)} = \frac{25 + 9}{10} = 3.4\Omega\). Voltage \(V_{AB} = I \times Z_{eq} = (5\angle 30^\circ) \times 3.4 = 17\angle 30^\circ~V\).

An AC source of RMS voltage 20V with internal impedance \(Z_s = (1 + 2j)\Omega\) feeds a load of impedance \(Z_L = (7 + 4j)\Omega\). The reactive power consumed by the load is

1. 8 VAR
2. 16 VAR
3. 28 VAR
4. 32 VAR

Solution

Total impedance \(Z_{total} = Z_s + Z_L = (1 + 2j) + (7 + 4j) = 8 + 6j~\Omega\). Magnitude of RMS current \(|I_{rms}| = \frac{|V_{rms}|}{|Z_{total}|} = \frac{20}{|8 + 6j|} = \frac{20}{\sqrt{8^2 + 6^2}} = \frac{20}{10} = 2A\). Reactive power consumed by the load \(Q_L = |I_{rms}|^2 \times X_L\), where \(X_L\) is the reactive part of \(Z_L\). \(X_L = 4\Omega\). \(Q_L = (2)^2 \times 4 = 16 \text{ VAR}\).