Solved Previous Year GATE Papers | Basic Electrical | 2012

Solution of GATE Previous Year 2012 Solved paper

Q1. A two phase load draws the following phase currents: \mathbf{I_{1}(t)=I_{m}sin(\omega t -\phi_{1})}, \mathbf{I_{2}(t)=I_{m}cos(\omega t -\phi_{2})}. These currents are balanced if \mathbf{\phi_{1}} is equal to\\\;\\\;\;\mathbf{(A)\;-\phi_{2}}\;\;\;\;\;\mathbf{(B)\;\phi_{2}\;\;\;\;\;} \mathbf{(C)\;(\frac{\pi}{2}-\phi_{2})\;\;\;\;\;} \mathbf{(D)\;(\frac{\phi}{2} + \phi_{2})}


\;I_{1}(t)=I_{m}Sin(\omega t - \phi_{1}) = I_{m}cos(90- (\omega t - \phi_{1}))\\\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\because cos(90-\theta)=sin(\theta)\\\;\\\;I_{1}(t)=I_{m}cos(\omega t - \phi_{1}-90)=I_{m}cos(\omega t -(\phi_{1}+90))\\\;\\I_{1}=I_{m}cos(\phi_{1}+90)

I_{2}=I_{m}cos(\omega t - \phi_{2}) = I_{m}cos \phi_{2}\\\;\\Now, in phasor form\\\;\\I_{1}=I_m\angle (90 + \phi_{1}) \\\;\\I_{2}=I_{2}\angle \phi_{2}

Currents will be balanced if \\I1 + I2 = 0 \\\;\\I_{m}\angle(90+\phi_{1})+I_{m}\angle \phi_{2} = 0

I_{m}cos(90+\phi_{1})+jI_{m}sin(90+\phi_{1}) + I_{m}cos\phi_{2}+jI_{m}sin \phi_{2} = 0\\\;\\I_{m}cos(90+\phi_{1})+I_{m}cos\phi_{2}+j(I_{m}sin(90+\phi_{1})+I_{m}sin\phi_{2}) =0\\\;\\I_{m}(cos(90+\phi_{1})+cos\phi_{2})+jI_{m}(sin(90+\phi_{1})+sin\phi_{2}) =0

cos(90+\phi_{1})+cos\phi_{2} = 0 \\\;\\cos(90+\phi_{1})=-cos\phi_{2}\\\;\\cos(90+\phi_{1})=cos(\pi+\phi_{2}) \\\;\\ 90 + \phi_{1} = \pi + \phi_{2}\\\;\\ \phi_{1}=\frac{\pi}{2}+ \phi_{2}

Q2. The impedance looking into the node 1 and 2 in the given circuit is\\\;\\\;\;\;\;\;\;\;\;\;

Solution: Simplify the diagram as show below\\\;\;\;\;\;

Now Apply voltage test source of 1A between the terminals “1” and “2” as shown below \\\;\;\;\;\;

We will slightly further modify the diagram and name the nodes as shown below.\\\;\;\;\;\;

Now Apply The Kirchhoff’s current law (KCL) at node “a”\\\;\;\;i_b + 99i_b + 1 = i_c\\\;\\\;\;\;100i_b + 1 = i_c\;\;\;\;\;\;\;\;\;\;\;(1)\\\;\\we know \\\;\\\;\;\;i_b = -\frac{V_{test}}{10K}\\\;\\\;\;\;i_c = \frac{V_{test}}{100}

Putting the values of Ib and Ic into equation (1)\\\;\\\;\;\;-\frac{V_{test}}{100}+1=\frac{V_{test}}{100}\\\;\\\;\;\;1 = \frac{V_{test}}{50}\\\;\\\;\;\;V_{test}= 50

Impedance or Resistance across the node “1” and “2” = \frac{V_{test}}{1}=\frac{50}{1}=50\Omega

Note: solved by using Thevenin theorem

Q3: In the circuit shown below, the current through the inductor is\\\;\\\;\;\;\;\;\\\;\\\;\;\;\;\;

Solution: Let current through 1\Omega is I2 as show in figure below\\\;\\\;\;\;\;\;

Applying KVL at node V1 \\\;\\1\angle 0 = I_{1}+I_{2}\\\;\\1\angle 0 = \frac{V_{1}-(-1\angle 0)}{j1}+\frac{V_{1}-(-1\angle 0)}{1}

j = V_{1}(1+j) +j + 1\angle 0 \\\;\\ V_{1}= -\frac{1}{1+j}

Now, Find I1 \\\;\\ I_{1} = \frac{V_{1}+1\angle 0}{j}\\Putting the Value of V1 in above equation \\\;\\I1 = \frac{1}{1+j}\\ Thus option “C” is correct.

Q4: The average power delivered to an impedance (4-j3)\Omega by a current 5cos(100\pi t+100)A is \\\;\\ (A) 44.2 W \;\;\; (B) 50W \;\;\; (C) 62.5 W \;\;\; (D) 125W

Solution: Given data \\\;\\Impedance , Z = 4 -j3 \\\;\\Current , I = 5cos(100\pi t+100)A

We Know

Average Power , P=I_{rms}^{2}R = I_{rms}V_{rms}cos\phi \;\;\;\;\;\;\;\;(1)

Impedance, Z = R +jX = 4 – j3 (Where R = Resistance, X = Reactance)\\\;\\Thus R = 4

Current , i = I_{m}cos(\omega t+ \theta) = 5cos(100\pi t + 100)\\\;\\Thus , Im = 5

We know, I_{rms}= \frac{I_{m}}{\sqrt{2}}= \frac{5}{\sqrt{2}}

Putting the Value of Irms and R in equation (1) \\\;\\Average Power, P = (\frac{5}{\sqrt{2}})^2 \times 4 = 50W

Q5: If VA – VB = 6V, then VC – VD is \\\;\\\\\;\\ (A) -5V \;\;\;\;\;(B) 2V \;\;\;\;\;(C) 3V \;\;\;\;\;(D) 6V

Solution: In the given circuit VAB = VA – VB = 6V\\\;\\ Current in the branch “AB” is I = \frac{V_{A}-V_{B}}{2}= \frac{6}{2}=3A

Consider The open port network as shown below

Current at one terminal is equal to current leaving the other terminal. Thus current in branch “AB” is equal to current in branch “DC”.

Total current in 1\Omega is 5A ( By KCL at node “D”)

Thus VC + 5 – VD = 0 \\\;\\VC – VD = – 5V (By Applying KVL to branch “CD” )

About AJAZ UL HAQ 134 Articles
AJAZ UL HAQ is an Assistant Electrical Engineer and has 10.5 Years of Experience in operation & maintenance, erection, Design of Transmission and Distribution Electrical System, Internal Electrification of Buildings, Electrical Substations. He has worked on Projects - Integrated Power development Scheme (IPDS), Prime Minister's Development Package (PMDP) and Re-structured Accelerated Power Development and Reforms Programme (RAPDRP).

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