GATE EE 2001


Question 1
In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor
A
is always zero
B
can never be greater than the input voltage
C
can be greater than the input voltage, however, it is 90^{\circ} out of phase with the input voltage
D
can be greater than the input voltage, and is in phase with the input voltage.
Electric Circuits   Resonance and Locus Diagrams
Question 1 Explanation: 
In a series RLC circuit, at resonance
V_L=jQV_{source} and V_c=-jQV_{source}
also for Q \gt 1, |V_c| \gt |V_{source}|
Hence, option (C) is correct.
Question 2
Two incandescent light bulbs of 40 W and 60 W rating are connected in series across the mains. Then
A
the bulbs together consume 100 W
B
the bulbs together consume 50W
C
the 60 W bulb glows brighter
D
the 40 W bulb glows brighter
Electric Circuits   Basics
Question 2 Explanation: 
\because \;\;P\propto \frac{1}{R}
Therefore , resistance of 40 W bulb \gt resistance of 60 W bulb.
For series connection, current through both the bulbs will be same P=I^2R (for series connection).
Power consumed by 40 W bulb \gtPower consumed by 60 W bulb.
Hencem the 40 W bulb glows brighter.


Question 3
A unit step voltage is applied at t = 0 to a series RL circuit with zero initial conditions.
A
It is possible for the current to be oscillatory.
B
The voltage across the resistor at t = 0^{+} is zero.
C
The energy stored in the inductor in the steady state is zero.
D
The resistor current eventually falls to zero.
Electric Circuits   Transients and Steady State Response
Question 3 Explanation: 
At t=0^+ inductor works as open circuit. Hence, complete source voltage drops across it and consequently, current through the resistor R is zero. Hence, voltage across the resistor at t=0^+ is zero. And further with time it rises accroding to V_R(t)=(1-e^{-Rt/L})u(t).

Question 4
Given two coupled inductors L_{1} and L_{2}, their mutual inductance M satisfies
A
M=\sqrt{L^{2}_{1}+L^{2}_{2}}
B
M \gt \frac{\left ( L_{1}+L_{2} \right )}{2}
C
M\gt \sqrt{L_{1}L_{2}}
D
M\leq \sqrt{L_{1}L_{2}}
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 4 Explanation: 
M=K\sqrt{L_1L_2}
where , K= coefficient of coupling
\because \; 0 \lt K \lt 1
\therefore \; M \leq \sqrt{L_1L_2}
Question 5
A passive 2-port network is in a steady-state. Compared to its input, the steady state output can never offer
A
higher voltage
B
lower impedance
C
greater power
D
better regulation
Electric Circuits   Two Port Network and Network Functions
Question 5 Explanation: 
For a passive two port network, output powe can never be grater than input power.


There are 5 questions to complete.