GATE Questions Archive - Grad Plus

# GATE Questions

## GATE EE 1998 Electrical Circuits Q 3

Viewed from the terminal A and B, the following circuit shown in figure can be reduced to an equivalent circuit of a single voltage source in series with a single resistor with the following parameter a) 10 volt source in …

## GATE EE 1997 Electrical Circuits Q 7

For the circuit shown in figure, The Norton equivalent source current value is _______ A and its resistance is _____ ohms Ans. 4.5 Ω Explanation: The given network is : By using source transformation,the circuit can be modified as, Again …

## GATE EE 1994 Electrical Circuits Q 7

Superposition principle is not applicable to a network containing time – varying resistors. (True/ False). Ans. False Explanation: The Superposition principle is applicable on both the time variant and time invariant resistors.So, the given statement is incorrect.

## GATE EE 2018 Electrical Circuits Q 1

The equivalent impedance Zeq for the infinity ladder circuit shown in the figure is a)j12Ω b)-j12Ω c)j13Ω d)13Ω Ans. (a) Explanation: Z1 = j9 Z2 = j5 – j1 = j4 Solving above equation we get, Zeq = j12

## GATE EE 2017 Electrical Circuits Q 2

The power supplied by the 25 V source in the figure shown below is _________W. Ans. P=250 W Explanation: By using KCL at node, We get I + 0.4I = 14 or I = 10 A The power supplied is …

## GATE EE 2017 Electrical Circuits Q 1

The equivalent resistance between the terminals A and B is ________ Ω. Ans. 3 Ω Explanation: Consider the following circuit daigram, After rearrangement We get,

## GATE EE 2016 Electrical Circuits Q 8

In the circuits shown below, the voltage and current sources are ideal. The voltage (Vout )across the current source, in volts is a) 0 b) 5 c) 10 d) 20 Ans. (d) Explanation:

## GATE EE 2016 Electrical Circuits Q 7

The voltage (V) and current (A) across a load are as follows. V(t)=100 sin(ωt) i(t) =10sin(ωt-60o)+2sin(3ωt) +5sin(5ωt). The average power consumed by the load, in W is________. Ans. 250 W Explanation: We know that the average power consumed by the …

## GATE EE 2016 Electrical Circuits Q 6

In the circuit shown below, the node voltage V is _________V. Ans. 11.42 V Explanation: Appling KCL to the given circuit at node A, We get 2 VA + VA -10 + 2 VA + 20I1 = 5 5 VA …

## GATE EE 2016 Electrical Circuits Q 5

A dc voltage with ripple is given by v(t) =[100 + 10sin(ωt)-5sin (3ωt)] volts. measurements of this voltage v(t),made by moving-coil and moving-iron voltmeters, show reading of v and v respectively. The value of V -V in volts is _________. …

## GATE EE 2016 Electrical Circuits Q 4

In the given circuits, the current supplied by the battery, in ampere is __________. Ans. I1 = 1/2 A Explanation: Given circuit is, By Applying KCL at node A, we get -I1 + I2 + I2 = 0 2I2 = …

## GATE EE 2016 Electrical Circuits Q 3

In the portion of a circuit shown \, if the heat generated in 5Ω resistances is 10 calories per second, then heat generated by the 4Ω resistances, in the calories per second is _________. Ans. Heat generated is 2 cal/sec …

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