GATE 2017 Electric Circuits Questions and Solutions

A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10\(\Omega\), 5\(\Omega\), 2\(\Omega\) are provided. Consider all possible permutations... The ratio of maximum to minimum values of the resistances (up to second decimal place) is _____.

Solution

The equivalent resistance is \(R_{eq} = R_A + (R_B || R_C) = R_A + \frac{R_B R_C}{R_B + R_C}\). Resistors: \{10\(\Omega\), 5\(\Omega\), 2\(\Omega\)\}.

Max \(R_{eq}\): Place the largest resistor (10\(\Omega\)) in position A. \(R_{eq,max} = 10 + (5 || 2) = 10 + \frac{5 \times 2}{5 + 2} = 10 + \frac{10}{7} = \frac{80}{7}\Omega\).

Min \(R_{eq}\): Place the smallest resistor (2\(\Omega\)) in position A. \(R_{eq,min} = 2 + (10 || 5) = 2 + \frac{10 \times 5}{10 + 5} = 2 + \frac{50}{15} = 2 + \frac{10}{3} = \frac{16}{3}\Omega\).

Ratio:

Consider the circuit shown in the figure. The Thevenin equivalent resistance (in \(\Omega\)) across P - Q is _____.

Solution

Apply test voltage \(V_{dc}\) across P-Q (10V source shorted). \(R_{TH} = V_{dc} / I_{dc}\). Let \(i_o\) be current down through vertical 1\(\Omega\) resistor. \(V_{dc} = i_o \times 1\Omega \Rightarrow i_o = V_{dc}\). KVL loop (1): \(-3i_o + i_o(1) + (i_o - I_{dc})(1) = 0 \Rightarrow -3i_o + i_o + i_o - I_{dc} = 0 \Rightarrow -i_o = I_{dc}\).

In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is

1. 1 nF
2. 1 \(\mu\)F
3. 1 mF
4. 10 mF

Solution

For MPT, \(Z_{load} = Z_{th}^*\). \(Z_{th} = R_S = 0.5\Omega\). So, \(Z_{load} = 0.5\Omega\). The load impedance is \(Z_{load} = j\omega L + (R || \frac{1}{j\omega C})\). Given \(R=1\Omega\), \(Z_{load} = j\omega L + \frac{1}{1 + j\omega C}\). Rationalizing:

We need \(Z_{load} = 0.5 + j0\). Real Part = 0.5: \(\frac{1}{1 + \omega^2 C^2} = 0.5 \Rightarrow \omega^2 C^2 = 1\). Imaginary Part = 0: \(\omega L - \frac{\omega C}{1 + \omega^2 C^2} = 0\). Substitute \(\omega^2 C^2 = 1\) into the imaginary part: \(\omega L - \frac{\omega C}{1 + 1} = 0 \Rightarrow C = 2L\). Given \(\omega = 100~rad/s\) and \(L = 5~mH = 0.005~H\).

For the network given in figure below, the Thevenin's voltage \(V_{ab}\) is

1. -1.5 V
2. -0.5 V
3. 0.5 V
4. 1.5 V

Solution

After the source transformations the network reduces to two branches meeting at a-b: a \(-30\)~V source behind 15\(\Omega\) and an 8~V source behind 5\(\Omega\). KCL at a, with b as reference, gives:

Multiply by 15:

For the given 2-port network, the value of transfer impedance \(Z_{21}\) in ohms is _____.

Solution

Number the \(\Delta\) nodes 1 and 2 (the upper terminals of port 1 and port 2) and 3 (the lower junction, which reaches the common terminal through the 2\(\Omega\) resistor). The \(\Delta\) arms are \(R_{12} = 2\Omega\), \(R_{13} = 4\Omega\), \(R_{23} = 2\Omega\), so \(\sum R = 8\Omega\). The equivalent Y arms are:

The network is now a T. The shunt arm is the path from node 3 to the common terminal, \(R_3 + 2\Omega = 1 + 2 = 3\Omega\), and the series arms are \(R_1 = 1\Omega\) and \(R_2 = 0.5\Omega\). For a T-network the transfer impedance is the shunt arm:

In the circuit shown below, the maximum power transferred to the resistor R is ____ W.

Solution

\(P_{max} = V_{th}^2 / (4R_{th})\). \(R_{th}\): Deactivate the sources and look into the R terminals. \(R_{th} = 5\Omega || 5\Omega = 2.5\Omega\).

\(V_{th}\): with R removed the remaining elements form a single loop. Taking the loop current \(I\) through the two 5\(\Omega\) resistors, KVL gives:

The open-circuit voltage at the R terminals is then:

The switch in the circuit, shown in the figure, was open for a long time and is closed at \(t=0\). The current \(i(t)\) (in ampere) at \(t = 0.5\) seconds is _____.

Solution

For \(t < 0\): \(i_L(0^{-}) = 5A\) (from current division). \(i(0^{+}) = i_L(0^{+}) = i_L(0^{-}) = 5A\). For \(t \to \infty\): Inductor is short. All 10A flows through the shorted inductor path. \(i(\infty) = 10A\). \(\tau\) (for \(t > 0\)): closing the switch short-circuits the 5\(\Omega\) in series with the inductor, so with the 10A source opened the inductor sees only the remaining 5\(\Omega\): \(R_{th} = 5\Omega\) and \(\tau = L / R_{th} = 2.5 / 5 = 0.5\)s.

With \(i(t) = i(\infty) + [i(0^{+}) - i(\infty)]e^{-t/\tau}\):

At \(t = 0.5\):