GATE 2024 Electric Circuits Questions and Solutions

In the given circuit, the current \(I_x\) (in mA) is _____.

Solution

This circuit is best solved using a super node around the dependent voltage source. Let \(V_1\) and \(V_2\) be the node voltages at the two ends of the dependent source. The super node equation (KCL) is:

The constraint equation (KVL) from the super node is:

We also know \(I_0 = \frac{V_1}{1~k\Omega} = \frac{V_1}{1000}\). Substituting \(I_0\):

Substitute (ii) into (1):

The current \(I_x\) is:

In the network shown below, maximum power is to be transferred to the load \(R_L\). The value of \(R_L\) (in \(\Omega\)) is _____.

Solution

For maximum power transfer, \(R_L = R_{TH}\). We find \(R_{TH}\) by applying a test source \(V_{dc}\) at the load terminals (with the 50V source shorted) and finding \(I_{dc}\). \(R_{TH} = V_{dc} / I_{dc}\).

Applying KVL in Loop (1) (bottom loop):

Following the PDF's derivation, we arrive at the relation: \(V_x = 1.5 I_{dc}\). Substitute this into (i):

Condition for MPT: \(R_L = R_{TH} = 2.5 \Omega\).

For the two port network shown below, the value of the Y21 parameter (in Siemens) is _____.

Solution

We first find the Z-parameters by applying KVL. Applying KVL in Loop (1):

Applying KVL in Loop (2):

Substitute \(V_1\) from (i):

The Z-matrix is \([Z] = \begin{bmatrix} 6/9 & 4/9 \\ -12/9 & 4/9 \end{bmatrix}\). The Y-matrix is \([Y] = [Z]^{-1}\).

The parameter \(Y_{21} = \frac{12}{8} = \frac{3}{2} = 1.5\) S.

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time \(t=0\). The time constant (in seconds) of the circuit for \(t>0\) is _____.

Solution

For \(t \ge 0\), the switch is closed. The time constant \(\tau\) for an R-L network is \(\tau = L/R_{TH}\). \(R_{TH}\) is the Thevenin resistance across the inductor (L) when all independent sources are deactivated (current source becomes open circuit). Looking into the terminals of the inductor: The 2\(\Omega\) resistor is in series with the parallel combination of the two 4\(\Omega\) resistors.

The time constant is:

Consider the circuit shown in the figure. The current I flowing through the 7\(\Omega\) resistor between P and Q (rounded off to one decimal place) is ____ A.

Solution

The circuit contains two parallel resistor pairs that can be simplified. Left side: \(3\Omega || 6\Omega = \frac{3 \times 6}{3+6} = \frac{18}{9} = 2\Omega\). Right side: \(2\Omega || 2\Omega = \frac{2 \times 2}{2+2} = \frac{4}{4} = 1\Omega\). The circuit reduces to a triangular loop.

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The 5A source current splits between two parallel paths: Path 1 (containing current I): 7\(\Omega\) resistor in series with the 2\(\Omega\) equivalent resistance. Total \(R_1 = 7+2=9\Omega\). Path 2: The 1\(\Omega\) equivalent resistance. By current division rule:

The Z-parameter matrix (with each entry in Ohms) of the network shown below is _____.

Solution

The network has a \(\Delta\)-network of 2\(\Omega\) resistors in the center. We convert this to a Y- network. Each resistor in the new Y-network \((R_1, R_2, R_3)\) is \(R_Y = \frac{R \times R}{3R} = \frac{2 \times 2}{3 \times 2} = 2/3~\Omega\). The circuit now becomes a T-network. The parameters of the equivalent T-network are: Left arm \(Z_a = 2\Omega + R_1 = 2 + 2/3 = 8/3~\Omega\). Right arm \(Z_c = 2\Omega + R_2 = 2 + 2/3 = 8/3~\Omega\). Shunt arm \(Z_b = R_3 = 2/3~\Omega\). The Z-parameters for this T-network are: \(Z_{11} = Z_a + Z_b = 8/3 + 2/3 = 10/3~\Omega\). \(Z_{22} = Z_c + Z_b = 8/3 + 2/3 = 10/3~\Omega\). \(Z_{12} = Z_{21} = Z_b = 2/3~\Omega\).