GATE 2026 Electric Circuits Practice Questions and Solutions

A network has a single node \(A\) above a reference node (ground). Three branches are connected between \(A\) and ground: (i) an ideal 12 V DC source in series with a \(4\,\Omega\) resistor, with the positive terminal of the source towards the resistor; (ii) a \(6\,\Omega\) resistor; and (iii) an ideal 3 A DC current source that pushes current into node \(A\). The Thevenin equivalent seen from terminals \(A\)-ground is

1. \(14.4\) V in series with \(2.4\,\Omega\)
2. \(7.2\) V in series with \(2.4\,\Omega\)
3. \(14.4\) V in series with \(10\,\Omega\)
4. \(7.2\) V in series with \(10\,\Omega\)

Solution

The open-circuit voltage is the node voltage \(V_A\). Apply KCL at \(A\), taking currents leaving the node as positive (Chapter 4):

Multiply through by 12:

Superposition confirms this: the 12 V source acting alone gives \(12\times 6/(4+6)=7.2\) V, and the 3 A source acting alone (12 V source shorted) gives \(3\times(4\parallel 6)=3\times 2.4=7.2\) V, summing to 14.4 V.

For \(R_{TH}\), suppress both independent sources: the voltage source becomes a short and the current source an open circuit. The \(4\,\Omega\) and \(6\,\Omega\) resistors are then in parallel between \(A\) and ground:

An ideal 20 V DC source is connected in series with a \(5\,\Omega\) resistor; this series combination is connected across a pair of terminals \(a\)-\(b\). A \(20\,\Omega\) resistor is also permanently connected across \(a\)-\(b\). A variable load resistor \(R_L\) is now connected across \(a\)-\(b\) and adjusted for maximum power transfer. The maximum power (in watt) delivered to \(R_L\) is _____.

Solution

Remove \(R_L\) and find the Thevenin equivalent at \(a\)-\(b\). With \(R_L\) open, the \(5\,\Omega\) and \(20\,\Omega\) resistors form a simple voltage divider across the 20 V source:

Suppressing the source (replacing it by a short) puts the two resistors in parallel:

The maximum power transfer theorem (Chapter 4) requires \(R_L=R_{TH}=4\,\Omega\), at which point the source voltage divides equally between \(R_{TH}\) and \(R_L\):

Check directly: \(I=16/(4+4)=2\) A, so \(P_L=I^{2}R_L=4\times 4=16\) W.

A \(100~\mu\)F capacitor is charged to 4 V. At \(t=0\) a switch connects it, through a \(2\,\text{k}\Omega\) resistor, to a 10 V DC source, with the polarity of the source aiding the existing capacitor voltage. The time (in seconds) at which the capacitor voltage reaches 8 V is _____ (round off to 2 decimal places).

Solution

This is a source-free-plus-forced first-order RC problem (Chapter 7). The time constant is

The initial value is \(v(0^{+})=4\) V (capacitor voltage cannot change instantaneously) and the final value is \(v(\infty)=10\) V, because at steady state no current flows through \(R\) and the capacitor sits at the source voltage. The general first-order response is

Set \(v(t)=8\) V and solve for \(t\):

The result is just over one time constant, as expected: at \(t=\tau=0.2\) s the voltage is \(10-6e^{-1}=7.79\) V, still short of 8 V.

A series RLC circuit has \(R=5\,\Omega\), \(L=0.5\) H and \(C=0.02\) F. The source is removed and the circuit is allowed to oscillate on its stored energy. The damped natural frequency of the resulting response is _____ rad/s (round off to 2 decimal places).

Solution

For the series RLC circuit the characteristic equation of the natural response is \(s^{2}+\dfrac{R}{L}s+\dfrac{1}{LC}=0\), giving the neper frequency \(\alpha\) and the undamped natural frequency \(\omega_0\) (Chapter 8):

Since \(\alpha<\omega_0\) the circuit is underdamped and the roots are complex:

For reference, critical damping would need \(R=2\sqrt{L/C}=2\sqrt{0.5/0.02}=2\times 5=10\,\Omega\); the given \(5\,\Omega\) is half of that, so oscillation is indeed expected.

A series combination of \(R=30\,\Omega\) and an inductor of reactance \(X_L=40\,\Omega\) is connected across the source \(v(t)=100\sqrt{2}\,\sin(\omega t)\) V. Taking the source voltage as the reference phasor, the steady-state current phasor (rms) is

1. \(2\angle -53.13^\circ\) A
2. \(2\angle +53.13^\circ\) A
3. \(2.83\angle -53.13^\circ\) A
4. \(1.41\angle -53.13^\circ\) A

Solution

The peak value of the source is \(100\sqrt{2}\) V, so its rms value is

The series impedance is (Chapter 9)

The negative angle is the expected lagging current for an inductive branch.

A single-phase load draws 10 kW at a power factor of 0.8 lagging from a 400 V, 50 Hz supply. A capacitor is connected in parallel with the load to raise the overall power factor to 0.95 lagging. The required capacitance is _____ \(\mu\)F (round off to 1 decimal place).

Solution

The capacitor supplies reactive power without changing the real power, so \(P\) stays at 10 kW while \(Q\) is reduced (Chapter 11).

Before correction, \(\cos\phi_1=0.8\), so \(\sin\phi_1=0.6\) and \(\tan\phi_1=0.75\):

After correction, \(\cos\phi_2=0.95\), so \(\sin\phi_2=\sqrt{1-0.9025}=0.31225\) and \(\tan\phi_2=0.31225/0.95=0.32868\):

The capacitor must supply the difference:

A capacitor across the 400 V supply delivers \(Q_C=V^{2}\omega C\), with \(\omega=2\pi(50)=314.16\) rad/s:

A balanced three-phase star-connected load with per-phase impedance \(Z_{ph}=(8+j6)\,\Omega\) is supplied from a balanced 400 V (line-to-line, rms), 50 Hz, three-phase source. The total real power drawn by the load is _____ kW (round off to 1 decimal place).

Solution

For a star connection the phase voltage is the line voltage divided by \(\sqrt{3}\), and the line current equals the phase current (Chapter 12):

The real power is dissipated only in the resistive part of each phase:

The line-quantity formula gives the same result:

Two magnetically coupled coils are connected in series. When they are connected series-aiding the measured equivalent inductance is 0.7 H, and when the connection of one coil is reversed to give series-opposing the measured equivalent inductance is 0.3 H. One of the coils has a self-inductance of 0.4 H. The coefficient of coupling between the coils is _____ (round off to 2 decimal places).

Solution

For two series-connected coupled coils the equivalent inductances are (Chapter 13)

Subtracting the two measurements isolates the mutual inductance:

Adding them gives the sum of the self-inductances:

With \(L_1=0.4\) H this gives \(L_2=0.5-0.4=0.1\) H. The coefficient of coupling is then

The value satisfies \(0\le k\le 1\), as it must for a passive pair of coils.

A two-port network has open-circuit impedance parameters \(Z_{11}=6\,\Omega\), \(Z_{12}=Z_{21}=3\,\Omega\) and \(Z_{22}=9\,\Omega\). Port 2 is terminated in a \(6\,\Omega\) resistor. The input impedance seen at port 1 is

1. \(5.4\,\Omega\)
2. \(6.0\,\Omega\)
3. \(4.5\,\Omega\)
4. \(3.6\,\Omega\)

Solution

The defining equations of the Z-parameters (Chapter 19), with both port currents taken as entering the network, are

The termination imposes \(V_2=-I_2Z_L\), because \(I_2\) enters the network while the load current leaves it. Substituting into the second equation:

Put this into the first equation and divide by \(I_1\):

A series RLC circuit with \(R=3\,\Omega\), \(L=1\) H and \(C=0.5\) F, with zero initial capacitor voltage and zero initial inductor current, is energised at \(t=0\) by a 10 V DC source. The peak value of the resulting current (in amperes) is _____ (round off to 2 decimal places).

Solution

Work in the \(s\)-domain with zero initial conditions (Chapter 16). The series impedance and the transformed source are

Partial fractions give \(\dfrac{10}{(s+1)(s+2)}=\dfrac{10}{s+1}-\dfrac{10}{s+2}\), so

The current starts at zero, decays to zero, and therefore has an interior maximum. Differentiate and set the derivative to zero:

At that instant \(e^{-t}=0.5\) and \(e^{-2t}=0.25\):

The response is overdamped, as expected from \(\alpha=R/2L=1.5\) against \(\omega_0=1/\sqrt{LC}=1.414\), so the current does not oscillate.